Please solve my question

[stevengoh911]

Given
x,y ∈ [0,2c]
f(x),g(y) ∈ [0,2c]
Show
| xy-f(x)+g(y)|≥ c²
exist

I don't really get this question.
Please help me

[krtxmrtz]

Given
x,y ∈ [0,2c]
f(x),g(y) ∈ [0,2c]
Show
| xy-f(x)+g(y)|≥ c²
exist

I don't really get this question.
Please help me

Welcome to the forums.

Is the question show that AT LEAST ONE PAIR of values (x,y) exists that verify the above inequality?

[jemidiah]

I think it's just asking you to show that, no matter what functions f and g are used, there MUST be a pair of x and y that satisfy the given inequality. If that's exactly what's written, the writer should word it better IMO.

Partial Solution:
Let x,y = 2c
xy = 4c^2

In the worst case, [-f(x) + g(y)] = [-2c + 0] = -2c

(I'm gonna assume c>0 since it's on the right side of the interval, though I'm pretty sure this will all work even if c is negative; c=0 is trivial)

|4c^2-2c| >= c^2 -> (i) 4c^2-2c >= c^2 or (ii) 4c^2-2c <= -c^2

(i) 4c^2-2c >= c^2 -> 3c^2-2c >= 0 -> 3c-2 >= 0 -> 3c >= 2 -> c >= 2/3
(ii) 4c^2-2c <= -c^2 -> 5c^2-2c <= 0 -> 5c-2 <= 0 -> 5c <= 2 -> c <= 2/5

So, for c >= 2/3 or c <= 2/5 [c>=0 as noted above], the statement is true for x,y = 2c, no matter what functions f and g are used.


Every attempt I make at getting into the rest of the range seems to end in a huge number of sub-cases, so I'm probably missing something.

That's my take on it.