New Problem,anyone Can?

:) SOLVE THE PROBLEM:
8^(sinx)^2+8^(cosx)^2=10+cos2y
Can u solve ?it reqires a little of intelligence

Phan, some of your problems contains two variables: x and y. I don't see how they can be solved. unless you want us to simplify. or maybe it is like last time: min = max lol

hmhm maybe this :

8^[1-cos^2(x)]+8^[cos^2(x)]=10+cos2y

=>

8
----------------- + 8^[cos^2(x)]=10+cos2y
8^[cos^2(x)]

let A=8^[cos^2(x)]

now:

8
----- + A =10+cos2y
A

8+ A^2
----------- = 10+cos2y
A


im going to prove this:


8+A^2
--------- <=9
A


=> A^2 - 9A +8 <=0

A1,2=[9+_sqrt(81-32)]/2

A1,2=8,1 (*)

+++ ;;;;;;;;;;;;;;;; ++++++++
------1---------------8-------------
;;;;;; ---------------- ;;;;;;;;;;;;;

we have now that
8+A^2
---------- <=9 (**)
A


and (|cos2y|<1) => 9<=(10-cos2y)<=11 (***)

using (**) & (***)

left side is equal to right only if both sides is equal to 9

now using (*) A1,2 = 1,8

(I) A=8^[cos^2(x)]=1
cos^2(x)=0 => {{{{ cos(x)=0 }}}}
10+cos2y=9 => {{{{ cos2y=1 }}}}

(II) A=8^[cos^2(x)]=8
cos^2(x)=1
10+cos2y=9
{{{{ cos(x)=1 }}}} or {{{{ cos(x)=-1 }}}}
{{{{ cos2y =1 }}}} {{{{ cos2y =1 }}}}

finaly:

(i) x=PI/2+k*PI ; k={0,1,........}
y=k*PI ; k={0,1,........}

(ii)x=2*k*PI ; k={0,1,........}
y=k*PI ; k={0,1,........}

(iii)x=PI+2*k*PI ; k={0,1,........}
y=k*PI ; k={0,1,........}


SEE YA

:D :D :D :D :D :o :o :o :o :o

yeah,well done!!!
now,wat do u want?,i'll give u

what do i want????????????? explain that

are u mathematic boy or something ,if these problems from ur school i mean maybe homework :D

:D :D :D :D :D :o :o :o :o :o

:mad: :mad: what ?u think this is my homework?never.i solved this problem and it just take me 5'.
:cool: its just for entertainment!

ok i didnt say that problem is difficult
and if u want to ill give u my math problems ,just say which part of math science.
have u ever been a competitior {of} mathematic?(in your country)
maybe olympiad?


how old are u?


:D :D :D :D :D :o :o :o :o :o

yeah,cause i think that's a exciting problem,so i wanna show you.if u have any math problem,show me ,i enjoy solving them.No,i'm just average at math,i'm not very good,yeah,maybe.
so you are?
i'm 15.5 old,n u?

well ok this is for the begining:

prove that 7 | [1/2(2+sqrt(3))^2002] + 1 where [1.6]=2 but [1.3] =1 etc.

but i think this is hard so i give u this too:

prove that there are {a lot of} i mean (where is my dictionary :D ) {maybe INFINITE lol } prime numbers ??

this is so easy if uve seen this problem till now



:D :D :D :D :D :o :o :o :o :o

so the bracket is rounding? [x] usually denotes the greatest integer less than or equal to x. it doesn't matter, it would be still 0, since 1/(really big number)=really small number
so [really small number] is 0
and what are you trying to prove? your question doesn't make sense to me.

by the way 7/((2+sqrt(3))^2002)+1=2360 (of course calculator)

maybe this is what ur trying to say.

as for the prime numbers,

assume there are a finite number (n) of primes

p1,p2,...,pn be the primes

let x=p1*p2*...*pn+1

x mod p1=x mod p2=...=x mod pn=1

therefore x is not divisible by any prime p1,p2,...,pn. therefore x is a prime. so proven by contradiction, there are an infinite amount of primes.

mmiill,
In your problem, do you mean prove that 7 divides evenly into [1/2(2+sqrt(3))^2002] + 1 ?

in 7 | [1/2(2+sqrt(3))^2002] + 1,wat is "l" ?can u write your question more clearly?please!! :)

if u want to compete with me,i suggest as follows:
u give me a problem and in return i 'll also give u one.The number of problem is 3,or 4...it's up to u Those who cant solve will be the loser after the winner show the loser the solution,ok? But the problem must be the maths of 10 or 11 grade?ok?

I am on.

just a note that these questions aren't really hard and can be done using pre-calculus methods. i am not trying to compete for whos the loser by the way. just trying to get something out of it by the end of the day.

1. find the smallest value for positive integers (n,k), other than n=1 and k=1
(1^2+2^2+...+n^2)/n=k^2

2. 3^{(1/2+log₃(cosx+sinx))}-2^{log₂(cosx-sinx)}=sqrt(2)

3. Let x, y > 0 be such that x³ + y³ <= x - y. Prove that x² + y² <= 1.

4. prove that for all positive real numbers a,b,c,

a³/bc+b³/ca+c³/ab>=a+b+c

determine when equality occurs.

[1/2(2+sqrt(3))^2002]+1 = 7*k is this true?????????

that is question and it isnt 2360 it's much more


and [1.6]=1 [1.3]=1 [1.9999]=1 in this case

lol ok. but LHS is a constant. so its 7k=c
k=c/7

one answer only

and what do you mean "is it true?" what's true?

and is it

[1/(2(2+sqrt(3))^2002)]+1 = 7*k

or actually

[((2+sqrt(3))^2002)/2]+1 = 7*k

ok,yesterday ,i cant access VBforums.now my solution is:
2.
<=>3^1/2(cosx+sinx)-(cosx-sinx)=2^1/2
<=>3^1/2*2^1/2cos(x-pi/4)+2^1/2sin(x-pi/4)=sqrt(2)
now,we can easily solve this equation (remember cosx+sinx>0and cosx-sinx>0)
3.for x, y > 0 ,x^3 + y^3 <= x - y if only if 0<x,y<1
let x=1/n;y=1/m for (n>2,m>1) and (n>1 and m>2)
=>(n^2+m^2)/(nm)^2>=2/nm >=1
=>x^2 + y^2 <= 1.
i havent seen 4,i'll show u later
i dont understand 1 ,tell me more clearly
i'll send u my problem later,now i must leave

:D :D

mmiill ,i havent seen your question clearly


:D :D :D

1.find the minimum of y=x^100-x^10+10
2.prove that for all positive numbers a,b,c,
a/(b+c)+b/(c+a)+c/(a+b)>=3/2
3.solve the equation: (2+sqrt(3))^x+(2-sqrt(3))^x=4x
4.find the max of f(x)=(sqrt(x^2-1))/2x^2:D

your solution for number 2 is out of line, at least thats what i think. i have no idea how you got the second line. and where does the cos(x-pi/4) come from? plz provide an answer in terms of x (Solve for x).

as for your number 3,

i agree with your third line, execept i don't see how it proves the original equation. and i don't quiet understand n^2+m^2<>1

what does <> mean? not equal? or do you mean n^2+m^2>1, since n>1 and m>1? and why did you make an assumption of n>1 and m>1? in the origial question, i said x>0, y>0 so x and y can be any real number.

for the first question, it asks you to find the smallest integer k, where k>1 and n is also an integer.

and (n,k) satisfies this equation:
(1²+2²+3²+...+n²)/n=k²

for the 4th question, it should read:

a³/(bc)+b³/(ca)+c³/(ab)>=a+b+c

I'll take a look at ur problems. tell me you can solve every one of them!!

{bu bre je...m mu mater cete da vam nacrtam !!!!!!!}


prove that 7 | [1/2*(2+sqrt(3))^2002]




show me the result of the first problem 1^2+2^2+......+n^2=k^2
if it exist ????



forth is very simple like 2,3 too ,but i can't show u becouse it's too long and i dont want desturbing my fingers. :D

and it would be very nice if u indicate my language mistakes . :p

and i think that it is too much 4 problems ,which is very simply,but too long for writeing ......
if u have hard ,interesting problem it is much better than 4 simply boring problems LOL


:D :D :D :D :D :o :o :o :o :o

stupid VB form destroyed my whole post :(

now i am going to explain question 1 in less detail:

taking the derivitive of y, we get when x=0 or x^90=1/10 a credical point occurs.

know we know that lim(x->oo)y=lim(x->-oo)y=oo, and since when x=0, x^100=x^10, but when x^90=1/10, x^100<x^10,

therefore the global minimum occurs when (x,y)=(1/10, 10^(-10/9)+10^(-1/9)+10), which i don't see how y can be simplified

mmill i don't think you can solve any of them. they may look easy but they actually aren't why don't you try it. if you can solve them (at least 3) i'll give you MUCH harder questions

i did number 2 before. the hard way is to take the common denominator and then everything cancels out. I am trying to remember an easier way to do it.

#4 looks like a pure calculus question: does it say:

f(x)=(sqrt(x^2-1)/2)*x^2 or

f(x)=sqrt(x^2-1)/(2x^2)?

ok ok ok
tell me which 3 problems i have to do and i ll do it ,and please dont saying anything agains me.
and what about my problem ???????

ok do u want :

1. i show u the result of ur 2-4 problems

or

2. guy from VIETNAM

sorry i didn't mean to be rude. anywayz you can do any problems you wish. i'd be interested in seeing your solution and how good you are in problem solving. i have like tons -- i mean tons of problems that are very hard. but the 4 problems i provided should be good enough for starters. they are very interesting might I add. if you want me to rank the difficulties of my problems, number 2 and 3 are much easier than 1 and 4. in fact, you can solve number 2 in 3 lines.

this is for now:

3. X^3+y^3<x-y

x^3-y^3<x^3+y^3<x-y

(x-y)(x^2+xy+y^2)<x-y

x^2+y^2+xy<1


x,y>0

=> x^2+y^2<1



4. a^3/bc +b^3/ac+c^3/ab =

a^4/abc + b^4/abc + c^4 /abc >a+b+c ???

a^4+b^4+c^4
-------------------- >a+b+c
abc


a^4+b^4>2sqrt(a^4b^4)=2a^2b^2

b^4+c^4>2sqrt(b^4c^4)=2b^2c^2

a^4+c^4>2sqrt(a^4c^4)=2a^2c^2

2(a^4+b^4+c^4)>2(a^2b^2+a^2c^2+c^2b^2)

a^4+b^4+c^4>a^2b^2+a^2c^2+c^2b^2

a^2b^2+a^2c^2+c^2b^2 >abc(a+b+c) ??????

2(a^2b^2+a^2c^2+c^2b^2) >2abc(a+b+c)

a^2b^2-2abac+a^2c^2 + a^2c^2-2acbc+c^2b^2+a^2b^2-2abbc+c^2b^2 >0 ????

(ab-ac)^2+(ac-cb)^2+(ab-bc)^2>0 and that is correct



[a^3/bc +b^3/ac+c^3/ab]*abc = a^4+b^4+c^4 > a^2b^2+a^2c^2+c^2b^2 > abc(a+b+c)

=>[a^3/bc +b^3/ac+c^3/ab]*abc > abc(a+b+c)

a^3/bc +b^3/ac+c^3/ab > a+b+c



and for now

2.(VIETNAM GUY)

a .. . . b . . . .c ..... . . . . . . . . 3
---- + ---- + ----- > 3 --------------------
b+c c+a a+b (b+c)/a+(a+c)/b+(a+b)/c


b/a+c/a+a/b+c/b+a/c+b/c>6 becouse

{

b/a+a/b >2

a^2+b^2
------------- >2 a^2-2ab+b^2>0 (a-b)^2>0 true
ab

}

now

. 9 . . . . . . . .. .. . ... .... . .. . .. . . . 9 .. .. . . . 3
------------------------------ > ---------------- = ------
(b+c)/a+(a+c)/b+(a+b)/c 6 2


OK i cant any more for now my arms are dead (and that is what im talking about)


SEE YA

:D :D :D :D :D :o :o :o :o :o

pretty good solutions.

for number 4, i'd like to show u an alternate solution:

WOLOG, a>=b>=c>=0,

((ab-c²)²+(a²-b²)²+ab(a-c)(b-c))/abc>=0

expand and and get:

a³/(bc)+b³/(ca)+c³/(ab)-a-b-c>=0

rearrange and get: a³/(bc)+b³/(ca)+c³/(ab)>=a+b+c

you seem to be very good. number 1 is pretty hard, try it. i'll find more harder problems if you like. give me a minute I am hungry and need food. :( :(

ok i am cooking and before the water is heated up i'll give you some harder problems:

you can take a look at this thread for some hard problems

http://www.vbforums.com/showthread.p...hreadid=189625

and i'll give you 3 really really REALLY HARD problems:

Write a formula which gives all solutions for:

1/x+1/y=1/z

x,y,z are integers

Find the whole part of the number:

4^(-1/3)+5^(-1/3)+6^(-1/3)+...+1000000^(-1/3)

and a little bit easier one:

factor a¹⁰+a⁵+1

4.
sqrt(x^2-1)/2x^2 = sqrt[(x^2-1)/4x^4]

1/2x^2 = t

sqrt(1/2t-t^2) this is max when 1/2t-t^2 is max


and that is 1/16 (i think u know why !!!)


and sqrt(1/16) =1/4 and that is the MAX of sqrt(x^2-1)/2x^2

could u tell me plz

first problem i dont understand it (becouse my english is bed)

end FACTOR a^10 ..... what is FACTOR ?????/

ok, first question is: given 1/x+1/y=1/z

write a formula that is always true for all solutions of x,y,z given that x,y,z are integers.

also, the last question wants you factor, namely writing the expression as the product of two algebraic expressions.

just like x^2+2x+1=(x+1)(x+1)

:D bugpodder
* u gave a wrong solution to 1,i'm waiting mmiill to solve.
and the second ,my explaination is:
cosx+sinx=sqrt(2) (sqrt(2)/2cosx+sqrt(2)/2sinx)=sqrt(2)(cosxcospi/4+sinxsinpi/4)
=sqrt(2)cos(x-pi/4);easy?the same to sin(x-pi/4).You can solve the equation like the way i show u above
*(n>2,m>1) or (n>1 and m>2).bet that u cant find any x,y>1 to satisfy
in PROBLEMS FOR JULY AND AUGUST ,161 is much easier than the forth,have u solved it yet?
mmiill,
u got a right solution to forth problem,but it's so long,can u get another shorter?
the second u solve so long,it 'll take me about 15' to read,please wait.......
yeah,u're very good at math.how old r u ?mind if u tell me?
wat about first and third one?
hey,why dont we three join the chat room to talk about math ,solving the hard problem?


I GOT THE FORTH BY MY WAY,ITS SO EASY,WANNA SEE IT?

:) mmiill,u seem to be so prideful to say that bugpodder's problem is 4 simply boring problems.but first ,finish my first and third one!!!
For the second,i dont really see your solution,can u write it more clearly?please!!
once u finish mine,i'll give the problem of 10 ,11 grade and if u can solve it,i'll
really recognize your talent.
********
bugpodder,
a^10+a^5+1 is easy to do.its as follow:
(a^5)^2+2*1/2a^5+1/4+3/4=(a^5+1/2)^2+3/4
an in PROBLEMS FOR JULY AND AUGUST,i dont completely see the geometry
problems(cause my eng. is not very good),if u show me the picture of those,it'll be easy for me to do
(160,163):D :) :D :D :D ;) ;) ;) ;)

Quote:

---

(a^5)^2+2*1/2a^5+1/4+3/4=(a^5+1/2)^2+3/4

---

thats not completely factored: there is a +3/4 trailing.
that is only PARTIALLY factored, so that is not the answer.


as the cos and sin one, i don't see where 3^(1/2) went, maybe you could post it step by step? if not i'll post my solution

yes, i did mess up a plus and minus in my answer: phan, the answer should be: (x,y)=(1/10, 10^(-10/9)-10^(-1/9)+10), i checked my answer for your question with my graphing calculator and it is right.

for 160, diagram is included. note i can't include everything in the diagram -- also concylic means it can be inscribed in a circle (a circle can be drawn such that it touches all the vertices of the quadrilateral).

by the way, the three questions that i gave later, i am positive that unless you are VERY GOOD you will not get it. thats why i like my first set of questions better, since you can do some of them, it at least get you confidence up. no point in giving something that no one could do (unless any of you can prove me wrong by doing one out of 3)

here is the image

3.

(2+sqrt(3))^x+(2-sqrt(3))^x=4x

2-sqrt(3)=1/(2+sqrt(3))

end let t=(2+sqrt(3))^x

we have:

t+1/t=4x

(t^2+1)/t=4x

t^2-4tx+1=0

t1,2=[4x+-sqrt(16x^2-4)]/2

t1,2=[4x+-2sqrt(4x^2-1)]/2

t1,2=2x+-sqrt(4x^2-1)


(2+sqrt(3))^x=2x+-sqrt(4x^2-1)

we can see that x=1 is the solution

also we can se that 2x>sqrt(4x^2-1)

2x+sqrt(4x^2-1)<4x

and also (2+sqrt(3))^x>2^x

and 2^x>4x x>=5

x=2,3,4 isnt correct

so x=1 is the solution

{but this is not the correct solution ,i think,becouse i {observe :D } just integers x>=1 ...............


when word is in brackets {} that means i dont have an idea what that word means!!!! :D

well it is very hard for me to write whole this text on my native language SERBIAN ,on ENGLISH it is much much more harder


an idea about chating is not so bad :D but i think :

look at this

CANADA SERBIA VIETNAM
12:00 19:00 02:00

i dont know something like that and i think that it is impossible to meet together :P
but we could try!!


SEE YA


:D :D :D :D :D :o :o :o :o :o

o my god !!!!

sorry for the my words about boring problems ,but thet is becouse my bad english i dont have an idea how i can say that .

i dont know what u dont understand about second problem,type me the line where mistake occures and ill get that

yes it is too long but that is becouse of {legibility} on my paper it is in a few lines.

and im wondering when u gonna give the solutions of yours problem

i cant sleep becouse of some problems .dont wait so much it could be bed for my health :D

and plz when u are talking about problem firs type which one not just first,second becouse thare are two second problems from u and our friend thanx


:D :D :D :D :D :o :o :o :o :o

well a10+a5+1

this is the stupid solution


[a^5+(-1+isqrt(3))/2][a^5-(-1-isqrt(3))/2] im just kiding

or more

[a^30-1][1/(a^10-1)] LOL

im very interested in problem about whole part of namber 4^(-1/3) + ..........

and

1^2+2^2+..........

can u give me some advice for that problems , and tell me that i dont have to use calculator for soloving......

VIETNAM :

your first problem about MIN value i think im gonna done that soon LOL (i mean i dont need solution for that becouse i think it is possibly to be done ) hahhaha i think that is the top of my ignorance of english





:D :D :D :D :D :o :o :o :o :o

ok, sorry,

for the

a^10+a^5+1, you are suppose to factor as the product of two polynomials, such as:

(a^3+2a^2-1)(a^7+1) -- which is not the answer of course

as for the 1^2+2^2+...+n^2, you must first realize, 1^2+2^2+...+n^2=n(n+1)(2n+1)

so now the question becomes:

(n+1)(2n+1)/6=k^2, find the smallest k such that k>1 and n,k are both integers.

i hope this makes easier.

for the whole number question:

(1+1/n)^2=1+2/n+1/n^2

(1+2/(3n))^3=1+2/n+4/(3n^2)+8/(27n^3)

so for every natureal number n,

(1+2/(3n))^3>(1+1/n)^2

thus

1+2/(3n)>(1+1/n)^(2/3)

multiply by n^(2/3) yields:

n^(2/3)+2/(3n(1/3))>(n+1)^(2/3)

and finally:

1/n^(1/3)>3/2((n+1)^(2/3)-n^(2/3))

this is the first half of the solution.

*************Bugpodder,
sorry, u 're wrong again,using derivitve is not good,i'm still waiting for mmiill....
for the second,i just show u the proof of cosx+sinx=sqrt(2)cos(x-pi/4)
=>3^1/2*2^1/2cos(x-pi/4)+2^1/2sin(x-pi/4)=sqrt(2)
<=>sqrt(6)cos(x-pi/4)+sqrt(2)sin(x-pi/4)=sqrt(2)
(acosx+bsinx=c,a^2+b^2>=c^2)
<=>(sqrt(3)/2)cos(x-pi/4)+(1/2)sin(x-pi/4)=(1/2)(divide both side by 2/sqrt(2))
cospi/6cos(x-pi/4)+sinpi/6sin(x-pi/4)=1/2
<=>cos(pi/6-x+pi/4)=cos(pi/3)
<=>cos(x-5pi/12)=cox(pi/3)
=>x-5pi/12=+-pi/3+k2pi
now u can get x (my finger need a rest).
*161 is so easy,160 is too now
*a^10+a^5+1 cause i dont understand what FACTOR is in VNese so i'm wrong but
now its ok...
********mmiill,
i'm rerally sorry for misunderstanding you.please dont be sad,we're still good friennd,ok?
the second problem is too difficult to read,mind if u write it more clearly?
i dont think the third is completely right,u didnt prove that x=1 is the root(we can see
x=1 all along)
wat about the first?
You're really very good at math,have u got any secret?

*************
WE ARE REALLY SO TIRED TO SOLVE SUCH HARD PROBLEM BUT
IF THERE'S A PRIZE ,IT WILL BE MORE INTERESTING,WONT IT?
I SUGGEST THAT :
EACH OF US WILL POST 3 PROBLEM,THE MAN WHO HAVE MOST RIGHT
SOLUTIONS ,WILL GET SOMETHING FROM THE OTHER,OK?
IF U AGREE ,TELL ME WAT THE PRIZE IS!!!
i must go to church now,see u later

:D hey,bugpodder, in 160,is I the incentre of the triangle and the center of inscribe circle?

lol i don't think i can ever mail anything to Viet Nam cuz i am out of $$$. well, let me tell you this, if you can solve these problems, the prize is of course, knowledge and experience, which enables you to do better in math. for me its practice for math contests (sorry Phan not online contests) i write every year. i mean its always good to learn some new techniques. lol i never knew cosx+sinx=sqrt(2)(cos(x-pi/4)) -- thats one hell of a nifty trick! and plus i can use some amusement in math lol! for one i like to see problems that uses trick like that!

ok, factor x^2+3x+2 would be come (x+1)(x+2)

both x+1 and x+2 have integral co-efficients. also there is no trailing term, so if you factor the above expression into (x+1)(x+1)+x+1, that would be wrong because you are adding something in the end.

LOL

prize ?? mmmmm i think it is interesting.

i dont know how it is in VIETNAM ,but i know how things works in CANADA.i have a friend in CANADA and he works like a PICA guy and he is payed about 1000$ for a mounth.
listen now in SERBIA it is catastrophe.a mounth pay in my country is about 100$ for people whit faculty.and that is very small pay. in CANADA it is about 15.000$ for people whit faculty.
and of course internet is so expensive.

im kiding about prize. :D


VIETNAM:
about our conflict it is OK :D . about third problem i know and i seid that it is not good,but i saw that when i finished writeing so i put that solution anyway.about first i havent looked till now but i will.
u said that u have to go to the church.it is very funy for me .what are u ortodox(hmmh i think u are not) or roman or muslim or maybe something else..(do u have a bais eyes like chinamen or ... :D)


CANADA:
thanks for a little help.



SEE YA


:D :D :D :D :D :p :p :p :p :p

a ... .. .. . . b . .. .. . .c ..... . . . . . . . . 3
------ + -------- + ---------- > 3 --------------------------------------
b+c . .. c+a . ... . a+b .. .... . . . (b+c)/a+(a+c)/b+(a+b)/c

this is becouse ( (a+b)/2 > 2/(1/a+1/b) )

aritmetic and ........... {disparity} i dont know on english

a1+a2+a3+......+an . . . .. . . n
-------------------------- > -------------------------
. .. . . . . .n . .. . . . . . . . ..1/a1+1/a2+.....+1/an

where 1,2,3, are indexs





b/a+c/a+a/b+c/b+a/c+b/c>6 becouse

{
(a-b)^2>0

=>
a^2-2ab+b^2>0

=>

a^2+b^2
------------- >2
ab

=>

b/a+a/b >2 (*)


(b/a)+[c/a]+(a/b)+{c/b}+[a/c]+{b/c}>6

becouse (*)

(b/a)+(a/b)+[c/a]+[a/c]+{c/b}+{b/c}> 2+2+2=6

}

now

. 9 . . . . . . . .. .. . ... .... . .. . .. . . . 9 .. .. . . . 3
------------------------------ > ---------------- = ------
(b+c)/a+(a+c)/b+(a+b)/c . . . . .6 .. . .. . . . .2

CANADA:
yes ive been looking for that formula since u gave that problem ,but i think u forgot to DIV whit 6.it is not just n(n+1)(2n+1) check it!

ok, here is the solution (i did forgot to divide by 6):

after you simiplify and rearrange, you get:

(4n+3)²=48n²+1

let x=4n+3

x²-48k²=1

so (x²-48k²)²=x⁴-96(kx)²+(48k²)²
=(x²+48k²)²-96(2xk)²

so since (7,1) satisfies x^2-48k^2=1
(1,7) (14,97)
then

(97,14) also satisfies the equation

but when x=97, n is not a integer

so similarly using the method above, we find:

(1351,195)

so n=(1351-3)/4=337 -- hehe told you its pretty hard

Quote:

---

a1+a2+a3+......+an . . . .. . . n
-------------------------- > -------------------------
. .. . . . . .n . .. . . . . . . . ..1/a1+1/a2+.....+1/an

---

this is called the AM-GM-HM inequality

The Arithmatic mean of a set of numbers is greater or equal to the Geometric Mean which is greater or equal the Harmonic Mean.

mm used AM>=HM

AM of a1,a2,a3,...,an=>(a1+a2+...+an)/n
GM => (a1*a2*...*an)^(1/n)
HM => n/(1/a1+1/a2+...+1/an)

let me write mm's solution more clearly

a/(b+c)+b/(c+a)+c/(a+b)>=3/2

By AM-HM

(a/(b+c)+b/(c+a)+c/(a+b))/3>=3/((b+c)/a+(c+a)/b+(a+b)/c)
a/(b+c)+b/(c+a)+c/(a+b)>=9/(b/a+c/a+c/b+a/b+a/c+b/c)

a/b+b/a=(a²+b²)/ab=(a²+b²-2ab+2ab)/ab=((a+b)²+2ab)/ab>2

so

a/(b+c)+b/(c+a)+c/(a+b)>=9/([b/a+a/b]+[c/a+a/c]+[c/b+b/c])
a/(b+c)+b/(c+a)+c/(a+b)>=9/6
a/(b+c)+b/(c+a)+c/(a+b)>=3/2

Ok ,thats just my idea
The mm'solution for 2 is so not too bad.What about the first?
hey,mmiill,y do u its funny to go to church?i'm roman catholic,
what 'bout u,muslim?

hahahahahahha


no im not muslim im ortodox ,just like russian.i dont like muslim becouse we had a lot of wars against muslim ,in bosnia,in kosovo,and u know for 11. september and for israel .............

im noe saying that there is not good one ,but most of they are fanatic ,and {suicider} and i hate them,becuse my country,s been in crisis becouse of them for 10 years,and still we have prombles in south SERBIA and KOSOVO..

olso i want to say im sorry for people who are muslims and who are good!!!!!
one of my best friend is muslim ,and there is no problem about that :D


about math problems i havent much time and i naveny seen for 2 days

CANADA:
are u shure about 1^2+2^2+........... ? that there is result!!!??? im just asking this :D







not too bed ??????????????????????????


it is same like CANADA guy said!!!!!!!!!!!









SEE YA

i gave u the solution, it is 4 posts up, mmiill.

mmiill,u are communist?cuz u r not catholic,hahahaha...
i hate the bad muslim ,the terrorist.
ignore it!!

hey,bugpodder,y dont u answer for my last post?u dont want me to help u solve 161? do u want the solution for 160?its very easy

lol it slipped my mind. but sure!

IM NOT COMMUNIST!!!!!!!

WHAT ARE U TALKING ABOUT ??????????????????




:mad: :mad: :mad: :mad: :mad:

:D i dont like communist.ok,forget it,lets get backto the maths,cant u solve the first? if not,i'll show u my sol

still npothing ihavent tried yey.but im sorry i cant tolk anymore whit u ,but nice to meet u.


SEE YA

what about whit sqrt 90 (1/10) for the firs problem PHAN.....