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I need to show that if $f: (a,b) \to \mathbb{ R }\text{ with}\;\; f''( x ) \geq 0$ for all $x \in (a,b)$, then $f\left( \frac{ x + y }{ 2 } \right) \leq \frac{ f( x ) + f( y ) }{ 2 }$.

I know that since $f''( x ) \geq 0$, then $f'(x)$ is monotone increasing. I'm not really sure where to go from here.

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Hint: Use this fact that if $f$ is differentiable in $I=(a,b)$ then it is concave upward on $I$ iff $$f(x)<\frac{x_2-x}{x_2-x_1}f(x_1)+\frac{x-x_1}{x_2-x_1}f(x_2)$$ for all $x_1,x,x_2$ in tne interval $I$ asuch that $x_1<x<x_2$. enter image description here

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Assume there would be any $x,y$ with $x\lt y$ and $f\left(\frac{x+y}{2}\right) \gt \frac{f(x)+f(y)}{2}$ then follows

$$f\left(\frac{x+y}{2}\right) \gt \frac{f(x)+f(y)}{2}\Rightarrow \frac{f\left(\frac{x+y}{2}\right)-f(x)}{\frac{x+y}{2}-x} \gt \frac{f(y)-f\left(\frac{x+y}{2}\right)}{y-\frac{x+y}{2}}$$

Applying mean valuee theorem on this twice will give you a value $w$ with $f''(w)\lt0$ which stands in conflict to $f''(x)\ge0\, \forall x$

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Very Nice point of view .+1 – Babak S. Dec 1 '12 at 14:01

Since the second derivative is positive, this implies that the $f$ is convex. So, just apply the finite form of Jensen's inequality with weights 1/2 to get the desired result.

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Yes, but how does one prove the first sentence? – Jesse Madnick Dec 1 '12 at 7:52
I guess that this all depends on how one defines convexity to start with. For differentiable functions, convexity could be defined as a property when the second derivative is positive. – Learner Dec 1 '12 at 8:13

You're right to observe that $f'(x)$ is monotone increasing, and in particular, this implies $f(x_1)\leq f(x_2$ for $x_1<x_2$, thus $f$ satisfies the definition of convexity. Now we can apply the finite form of Jensen's inequality with $a_1=a_2=\frac{1}{2}$, and we get $$f\left(\frac{x_1+x_2}{2}\right)\leq\frac{f(x)+f(y)}{2},$$ as desired. I hope this helps!

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Since $f''(x) \geq 0$, $f'(x)$ is monotone increasing (as you already observed). Let $x, y \in (a, b)$ and WLOG $x < y$. Therefore, $x < \frac{x + y}{2} < y$. Now, applying MVT in $\left(x, \frac{x+y}{2}\right)$ and $\left(\frac{x + y}{2}, y\right)$, we get $$ f'(c_1) = \frac{f\left(\frac{x + y}{2}\right) - f(x)}{\frac{y - x}{2}}\\ f'(c_2) = \frac{f(y) - f\left(\frac{x + y}{2}\right)}{\frac{y - x}{2}} $$ for some $c_1 \in \left(x, \frac{x+y}{2}\right)$ and some $c_2 \in \left(\frac{x + y}{2}, y\right)$.

Now $c_1 \leq c_2$, therefore $f'(c_1) \leq f'(c_2)$. This implies, $$ \frac{f\left(\frac{x + y}{2}\right) - f(x)}{\frac{y - x}{2}} \leq \frac{f(y) - f\left(\frac{x + y}{2}\right)}{\frac{y - x}{2}} \\ f\left(\frac{x + y}{2}\right) \leq \frac{f(x) + f(y)}{2} $$

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I would say: USE THE MEAN VALUE THEOREM. (Yes, I'm shouting.)

First, if $a<b$ then $f'(a) \le f'(b)$, since $(f'(b)-f'(a))/(b-a)$ is equal to $f''$ at some point between $a$ and $b$.

Now let $a<b<c$, note that $(f(b)-f(a))/(b-a) \le (f(c)-f(b))/(c-b)$ since the first difference quotient is $f'$ at some point between $a$ and $b$ and the second difference quotent is $f'$ at some point between $b$ and $c$. Rearrange this inequality to get $$ f(b) \le \frac{c-b}{c-a}f(a)+\frac{b-a}{c-a}f(c) $$ which is more general than the inequality in the question.

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