<s>[INST] Prove that $3(a^3+b^3+c^3) > (a+b+c)(a^2+b^2+c^2)$ Let $a,b,c$ be positive, not equal. Prove that $3(a^3+b^3+c^3) > (a+b+c)(a^2+b^2+c^2)$.
I know the proof by subtracting LHS by RHS and then doing some arrangement.
But isn't there any inequality which can be used in $(1+1+1)(a^3+b^3+c^3) >(a+b+c)(a^2+b^2+c^2)$, or an easy proof?
 [/INST] The result is also a consequence of the Chebyshev (sum) Inequality.
This is a quite useful result, not the least for contest problems!  It can be proved by using the Rearrangement Inequality, which is also useful to know. For completeness, we state the full result, though only half of it is needed here.
Theorem: (Chebyshev Inequality) Suppose that $a_1 \le a_2 \le \cdots \le a_n$ and $b_1 \le b_2 \le \cdots \le b_n$.
Let $m=a_1b_n + a_2b_{n-1} +\cdots + a_nb_1$ and $M=a_1b_1+a_2b_2+\cdots + a_nb_n$. Then
$$nm \le (a_1+a_2+\cdots+a_n)(b_1+b_2+\cdots +b_n) \le nM$$
with equality only happening when all the $a_i$ are equal or all the $b_i$ are equal.
To apply the result to our problem, we can without loss of generality assume that
$a \le b \le c$.  Let $a_1=a$, $a_2=b$, $a_3=c$, $b_1=a^2$, $b_2=b^2$, and $b_3=c^2$.  
Then the conditions of Chebyshev's Inequality are met, and $M=a^3+b^3+c^3$.  We can then read off our inequality from the Chebyshev Inequality.  
Note that the Chebyshev Inequality yields an immediate generalization  of the $3$ variable inequality of the question to $n$ variables.
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