# Prove that for all positive real numbers $a,b,$ and $c$ we have $a^5+b^5+c^5 \geq a^3bc+ab^3c+abc^3$

Prove that for all positive real numbers $a,b,$ and $c$ we have $a^5+b^5+c^5 \geq a^3bc+ab^3c+abc^3$.

This question reminds me of rearrangement, but I can't really find two sequences that fit. Maybe there is a way using the triangle inequality, but I am unsure.

We have $$\frac{3}{5}a^5 + \frac{1}{5}b^5 + \frac{1}{5}c^5 \geq a^3bc$$

by weighted AM-GM. Permuting the variables and adding gives the result.

(Incidentally, this approach can generally be used to prove Muirhead's inequality.)

Since $f(a,b,c)=LHS-RHS$ is homogeneous, we can asumme $abc=1$. Thus, inequalitie is equivalent to $a^5+b^5+c^5\ge a^2+b^2+c^2$ subject to $abc=1$.

Now, if $a\ge b\ge c$, by Tchebyshev's inequality,

$$\begin{eqnarray}\dfrac{a^5+b^5+c^5}{3}&=&\dfrac{a^3a^2+b^3b^2+c^3c^2}{3}\\&\ge&\dfrac{a^3+b^3+c^3}{3}\dfrac{a^2+b^2+c^2}{3}\\&\ge&\sqrt[3]{a^3b^3c^3}\dfrac{a^2+b^2+c^2}{3}\\&=&\dfrac{a^2+b^2+c^2}{3}\end{eqnarray}$$

and the result follows.

Tchebyshev Inequality: If $a_1\le a_2\le...\le a_n$ and $b_1\le b_2\le...\le b_n$, then $$\dfrac{a_1b_1+a_2b_2+\cdots+a_nb_n}{n}\ge\dfrac{a_1+a_2+\cdots+a_n}{n}\dfrac{b_1+b_2+\cdots+b_n}{n}.$$

Proof Applying rearrangement inequality:

$$\begin{array}{rcl} a1b_1+a_2b_2+...+a_nb_n&=&a_1b_1+a_2b_2+...+a_nb_n\\ a1b_1+a_2b_2+...+a_nb_n&\ge&a_1b_2+a_2b_3+...+a_nb_1\\ a1b_1+a_2b_2+...+a_nb_n&\ge&a_1b_3+a_2b_4+...+a_nb_2\\ \vdots&\vdots&\vdots\\ a1b_1+a_2b_2+...+a_nb_n&\ge&a_1b_n+a_2b_1+...+a_nb_{n-1}\\ \end{array}$$

Adding all the expressions give us the desired result. Note that equality is done iff $a_1=a_2=...=a_n$ or $b_1=b_2=...=b_n$

PD: I don't know if this Tchebyshev is the same of Statistics Theory

• I don't get the Tclebyshev's inequality part. – Puzzled417 Jan 17 '16 at 3:40
• Ok. I'll put it in my answer – sinbadh Jan 17 '16 at 3:44

Since $(5,0,0)$ majorizes $(3,1,1)$, this follows by Muirhead's Inequality.

• your reasoning does not help here because also $(3,2,0)\succ(3,1,1)$, but $a^3b^2+b^3c^2+c^3a^2\geq a^3bc+b^3ac+c^3ab$ is wrong! – Michael Rozenberg Jan 17 '16 at 14:10
• You have applied Muirhead incorrectly: by $(3,2,0)$ you must mean the fully symmetrized sum $a^3b^2+a^2b^3+\cdots$, not just the cyclically symmetrized sum you have written down. For the pairs $(5,0,0)$ and $(3,1,1)$ there is no distinction between the fully symmetrized and the cyclically symmetrized (up to a factor of $2$ on both sides). Does this clear your confusion? – pre-kidney Jan 17 '16 at 23:45
• now it's clear. – Michael Rozenberg Jan 18 '16 at 12:30

Also, we can use SOS: $$\sum_{cyc}(a^5-a^3bc)=\sum_{cyc}a^3(a^2-bc)=$$ $$=\frac{1}{2}\sum_{cyc}a^3((a-b)(a+c)-(c-a)(a+b))=$$ $$=\frac{1}{2}\sum_{cyc}(a-b)(a^3(a+c)-b^3(b+c))=$$ $$=\frac{1}{2}\sum_{cyc}(a-b)^2((a+b)(a^2+b^2)+c(a^2+ab+b^2))\geq0.$$