Questions tagged [karamata-inequality]

We'll say that $$(a_1,a_2,...,a_n)\succ(b_1,b_2,...,b_n)$$ iff $$a_1\geq a_2\geq...\geq a_n,$$ $$b_1\geq b_2\geq...\geq b_n$$ and $$a_1\geq b_1,$$ $$a_1+a_2\geq b_1+b_2,$$ $$\cdot$$ $$\cdot$$ $$\cdot$$ $$a_1+a_2+...+a_n=b_1+b_2+...+b_n.$$

The Karamata's inequality says:

• Let $(a_1,a_2,...,a_n)\succ(b_1,b_2,...b_n)$ and $f$ is a convex function. Prove that: $$f(a_1)+f(a_2)+...+f(a_n)\geq f(b_1)+f(b_1)+...+f(b_n).$$

• Let $(a_1,a_2,...,a_n)\succ(b_1,b_2,...b_2)$ and $f$ is a concave function. Prove that: $$f(a_1)+f(a_2)+...+f(a_n)\leq f(b_1)+f(b_1)+...+f(b_n).$$

A proof for the convex function.

Induction respect to $n$.

For $n=1$ and $n=2$ it's obvious.

For example, for $n=2$ let $A_1(a_1,f(a_1)),$ $A_2(a_2,f(a_2)),$

$B_1(b_1,f(b_1)),$ $B_2(b_2,f(b_2))$,

$M\left(\frac{a_1+a_2}{2},\frac{f(a_1)+f(a_2)}{2}\right)$ and $M\left(\frac{b_1+b_2}{2},\frac{f(b_1)+f(b_2)}{2}\right)$.

Since $f$ is a convex function, the segment $A_1A_2$ is placed above the segment $B_1B_2$, which gives $M$ is placed above $N$, which says $f(a_1)+f(a_2)\geq f(b_1)+f(b_2).$

Now, let $f(a_1)+f(a_2)+...+f(a_k)\geq f(b_1)+f(b_1)+...+f(b_k)$ for $(a_1,a_2,...,a_k)\succ(b_1,b_2,...b_k)$ for any $k\leq n-1$ and $n\geq2.$

Let $b_1+b_n=constant$ and $b_1$ increases.

Let for $b_1'\geq b_1$ in the first time happens: $$a_1+a_2+...+a_l=b_1'+b_2+...+b_l.$$ Thus, $$a_1\geq b_1',$$ $$a_1+a_2\geq b_1'+b_2,$$ $$\cdot$$ $$\cdot$$ $$\cdot$$ $$a_1+a_2+...+a_l=b_1'+b_2+...+b_l,$$ $$a_1+a_2+...+a_{l+1}\geq b_1'+b_2+...+b_{l+1},$$ $$\cdot$$ $$\cdot$$ $$\cdot$$ $$a_1+a_2+...+a_n=b_1'+b_2+...+b_n'$$ or $$a_1\geq b_1',$$ $$a_1+a_2\geq b_1'+b_2,$$ $$\cdot$$ $$\cdot$$ $$\cdot$$ $$a_1+a_2+...+a_l=b_1'+b_2+...+b_l,$$ $$a_{l+1}\geq b_{l+1},$$ $$\cdot$$ $$\cdot$$ $$\cdot$$ $$a_{l+1}+a_{l+2}+...+a_n=b_{l+1}+b_{l+2}+...+b_n',$$ which by the assumption of the induction gives: $$f(a_1)+f(a_2)+...+f(a_l)\geq f(b_1')+f(b_2)+...+f(b_l)$$ and $$f(a_{l+1})+f(a_{l+2})+...+f(a_n)\geq f(b_{l+1})+f(b_{l+2})+...+f(b_n'),$$ which gives $$f(a_1)+f(a_2)+...+f(a_n)\geq f(b_1')+f(b_2)+...+f(b_n')$$ and since by Karamata for $n=2$ we have $$f(b_1')+f(b_n')\geq f(b_1)+f(b_n),$$ we are done!