# Proving Jensen's inequality.

I've been wondering about this problem for some time. This is in a way related to the proof of Jensen's inequality. Say we have a convex function $f$. Can we prove $$\frac{f(a)+f(b)+f(c)}{3}\geq f(\frac{a+b+c}{3})$$ using only the fact that $$\frac{f(a)+f(b)}{2}\geq f(\frac{a+b}{2})$$

The proof that I know for Jensen's inequality generalises the problem to a form I wouldn't have guessed on my own. The proof is given here.

• Yes. Prove the case of four variables using the 2-variable case twice, and substitute $d = (a+b+c)/3$ for the fourth variable. – user27126 Aug 7 '13 at 7:18
• Convexity is usually defined a bit more general than your last inequality. Then write $$\frac{f(a)+f(b)+f(c)}{3} = \frac{2}{3} \frac{f(a)+f(b)}{2}+\frac{f(c)}{3}.$$ – WimC Aug 7 '13 at 7:19
• @WimC- I had tried your method before. It leads to infinite iterations without a solution. – fierydemon Aug 7 '13 at 7:48
• I did not hint at iteration but at the property that $$t f(x) + (1-t) f(y) \geq f(t x +(1-t)y)$$ for all $t\in[0,1]$. Then use $t=\frac{1}{2}$ and $t=\frac{2}{3}$ to derive an inequality for the RHS. Or do you specifically request to use only $t=\frac{1}{2}$? – WimC Aug 7 '13 at 8:29

Just as Sanchez said, you can in fact prove that $$\frac{f(x_1)+f(x_2)+...+f(x_n)}{n} \geq f(\frac{x_1+x_2+...x_n}{n})$$
First prove this for $n=2^k$ by induction on $k$. The induction step is just applying the induction hypothesis twice.
Second, knowing the proposition is true for $n=2^k$ prove it for each $i<2^k$. To do this apply the $2^k$ inequality for $x_1,...,x_i$ plus additional $2^k-i$ copies of $$y=\frac{x_1+x_2+...x_i}{i}$$ After easy transformations this will turn equivalent to the proposition for $n=2^k$