If $f(x)$ is a concave upward function in the interval $x_1,x_2$, prove that $f\left(\frac{a+b}{2}\right)\leq \frac{f(a)+f(b)}{2},x_1\leq a,b\leq x_2$
The slope of a concave upward function increase continuously, i.e its second derivative is positive. For simplicity's sake I'll assume $f(b)=b$ and draw a line $y=x$. $m=a-f(a)$ and $n=(a+b)/2-f((a+b)/2)$
$\frac{f(a)+f(b)}{2}=(a+b)/2-m/2$
$f\left(\frac{a+b}{2}\right)=(a+b)/2-n$
$m>n$, but here we are taking $m/2$ so its not of much help :(. How can I proceed further?
Best Answer
Let $$ F(x) = \frac{f(a) + f(x)}{2} - f\left(\frac{a+x}{2}\right)\;\;, $$ wherefrom $$ F'(x) = \frac{1}{2}\left[f'(x) - f'\left( \frac{a+x}{2} \right) \right]\;\;. $$ Evidently, $~F(a) = 0~$ and $~F'(a) = 0~$.
For $x >a$, we have $F'(x)>0$, provided $f$ is concave-up and both $x$ and $a+x$ reside within the permitted domain. Thence, for the said values of $x$, we obtain $F(x)>0$.
The case of $X<0$ is treated similarly.