More Algebra

[sparky69er]

1)Show for all real numbers a, b > 0, that

(a + b) > or equal to (2)( the square root of ab)

2) show that (a + b)(1/a + 1/b) = [(a + b)^2] / ab

3) Deduce that (a + b) (1/a + 1/b) > or equal to 4

[opus]

I talked to your teacher, he wants You to solve those!

[System_Error]

seriously, this forum needs a moderator and rules against blatant 'do my homework for me' posts.

[sevenhalo]

1)Show for all real numbers a, b > 0, that

(a + b) > or equal to (2)( the square root of ab)

2) show that (a + b)(1/a + 1/b) = [(a + b)^2] / ab

3) Deduce that (a + b) (1/a + 1/b) > or equal to 4

1 - ((A + B)2/(A + B))

2 - (A + B)/(a / 2) + (A + B)/(b / 2) = (A2 + B2)

3 - 2 pretty much proves it.

And if this isn't right, QC (Quality Control) will send you an email after a client complains.

[Abbas Haider]

Problem 2:

(a+b)(1/a+1/b)=1+a/b+b/a+1
=a/b+b/a+2
Multiplying the eq by ab/ab
gives (a^2+b^2+2ab)/ab
= (a+b)^2/ab

[Abbas Haider]

Problem 1:

since
(a+b)^2=a^2+b^2+2ab
Since
a^2+b^2>=2ab for a,b>0
Therefore
Forcing a^2+b^2=2ab
results in inequality (a+b)^2>2ab+2ab
but as a^2+b^2 can be equal to 2ab
so (a+b)^2=2ab+2ab
it implies that
(a+b)^2>=2ab+2ab
= (a+b)^2>=4ab
Sq. root both the sides
a+b>=2*Sqrt(ab)

[Abbas Haider]

I have solved problem 1 for you,
How about you solve this assumption I take
that
a^2+b^2 >= 2ab

Cmon it would be fun

[Abbas Haider]

Problem 3:
(a+b)(1/a+1/b)>=4
1+a/b+b/a+1>=4
a/b+b/a+2>=4
(a^2+b^2)/ab+2>=4
(a^2+b^2+2ab)/ab>=4
a^2+b^2+2ab>=4ab
a^2+b^2>=4ab-2ab
a^2+b^2>=2ab (prooved)

Since I take the assumption a^2+b^2>=2ab. I am waiting for you to prove that a^2+b^2>=2ab
I have solved it by putting values to it.