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=Could anyone help me show that:

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


$$ f(ax + (1-a)y) \geq af(x) + (1-a)f(y) $$ 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.


I managed to get:

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


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Did you plot the function and look at the graph? (It is concave, not convex), so you seem to be doing everything right. – t.b. May 4 '12 at 15:49
Sorry, I meant concave... It does not seem to work. I know that f(ax + (1-a)y) >= af(x) + (1-a)f(y) – Bober02 May 4 '12 at 15:54
According to the most common definition of convex, you are right. Ah, just saw the change. Sure it is concave, use first year calculus, second derivative is negative. – André Nicolas May 4 '12 at 15:54
Well, I meant if anyone could show it using the formula in my comment I would really appreciate it.... – Bober02 May 4 '12 at 15:56
I modified my question to include my desired formula – Bober02 May 4 '12 at 15:59
up vote 2 down vote accepted


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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OK, what about the Right hand side? the right hand side has ax^2, not a^2x^2 – Bober02 May 4 '12 at 16:17
got it ! thanks a lot :) – Bober02 May 4 '12 at 16:22

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? – Bober02 May 4 '12 at 15:57

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