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=Could anyone help me show that:

$ f(x) = -x^2 + 2x $

using

$ f(ax + (1-a)y) \geq af(x) + (1-a)f(y) $ is CONCAVE in $(0,1)$? I am trying to solve it by directly substituting to the general theorem but I sem to prove just the opposite.

Update:

I managed to get:

$ -(ax + (1-a)y)^2 \geq -ax^2 - y^2 + ay^2 $

Anyone?

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    I modified my question to include my desired formula2012-05-04

2 Answers 2

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Hint

Recall the arithmetic-mean-geometric mean inequality which gives

$ 2ab \leq a^2 + b^2 $

(which you can derive from the fact that $(a-b)^2 \geq 0$ for real numbers).

Apply it to

$ [ax + (1-a)y]^2 = (ax)^2 + (1-a)^2y^2 + 2a(1-a) xy $

in the form of

$ 2xy \leq x^2 + y^2 $

will lead you to the result you want.

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    got it ! thanks a lot :)2012-05-04
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Basic calculus is the easy way: the second derivative is negative.

If you want to use the formal definition, there is a bit less hassle if you note that $-x^2+2x=-(x-1)^2+1$. Now make the change of variable $w=x-1$, just a shift, does not affect the geometry. So we are looking at the function $-w^2+1$. Pull it down by $1$, does not affect the geometry. We are trying to prove that $-w^2$ is concave. I don't like negative numbers much, flip sign. So we want to prove that $w^2$ is convex.

Let $0 \le t\le 1$. We want to show that $(tx+(1-t)y)^2 \le tx^2+(1-t)y^2$. Expand the left-hand side. We want to show that $tx^2+(1-t)y^2-t^2x^2-2t(1-t)xy-(1-t)^2y^2 \ge 0$.

A bit of algebra reduces the above expression to $t(1-t)(x^2-2xy+y^2).$ It is not difficult to show this is non-negative!

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    I know, would you attempt solving it using the frmula in my comment?2012-05-04