Prove $a^3+b^3+c^3 + 3abc \ge a^2(b+c)+b^2(a+c)+c^2(a+b)$ for $a,b,c >0$

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Problem:
Prove $a^3+b^3+c^3 + 3abc \ge a^2(b+c)+b^2(a+c)+c^2(a+b)$, for $ a,b,c >0$

I've been playing around with this inequality for a while but kept running into dead ends. I've tried AM-GM inequality, tried to establish bounds, ect...
The one fact that I found probably useful is that: $(a+b+c)(a^2+b^2+c^2)=a^3+b^3+c^3 + a^2(b+c)+b^2(a+c)+c^2(a+b)$. But still couldn't make good use of it.

Does this answer your question? You are given the sides of a triangle $a,b,c$ Prove that: $a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$ (Sweden 1950) - found using an Approach0 search. — John Omielan, Jul 25 '22 at 05:23
FYI, there's also the one closed as a duplicate of the one above, i.e., Prove that $a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c) \leq 3abc.$, plus Prove inequality in a triangle, where its answer explicitly states your inequality is just a rewritten version. In addition, there's the related AoPS threads Hard or Trivial?, can you solve this problem? and Plis enlighten me. — John Omielan, Jul 25 '22 at 05:28

score 0 · Answer 1 · answered Jul 25 '22 at 05:10

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Hint: Use Schur's inequality for the case $r = 1$. You essentially prove Schur's inequality special case. You might look up Wikipedia about this particular powerful inequality. I highly doubt it that you can prove it using AM-GM or CS inequality.

answered Jul 25 '22 at 05:10

Wang YeFei

6,390

Prove $a^3+b^3+c^3 + 3abc \ge a^2(b+c)+b^2(a+c)+c^2(a+b)$ for $a,b,c >0$

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