<s>[INST] 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.
 [/INST] 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
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