I am trying to prove the inequality 3*(a^3+b^3+c^3) > (a+b+c)*(a^2+b^2+c^2)
where a,b,c are positive and unequal.
RHS = a^3+b^3+c^3+ab*(a+b)+bc*(b+c)+ac*(a+c)
so the inequality reduces to proving
2*( a^3+b^3+c^3) > ab*(a+b)+bc*(b+c)+ac*(a+c)
This is where I got stuck , how to prove this ? , please help.
Note that $$a^3+b^3>ab(a+b)$$ since we have $$(a+b)(a^2-ab+b^2)>ab(a+b)$$
Assume WLOG that $a\le b\le c$. Then $a^2\le b^2\le c^2$.
Hence, it follows from Chebychev’s inequality that $$3\cdot (a\cdot a^2 + b\cdot b^2 + c\cdot c^2)\ge (a+b+c)\cdot(a^2+b^2+c^2)$$
Note: if $a_1\le a_2\le\cdots\le a_n$ and $b_1\le b_2\le\cdots\le b_n$, then Chebychev’s inequality says that $$n\cdot\sum_{i=1}^na_ib_i\ge\sum_{i=1}^na_i\cdot\sum_{i=1}^nb_i$$
