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This is a question derived from another exercise problem for high school math contests. I hope this new question is also interesting.

If $a>0$, $b>0$, and $c>0$ with $abc=1$, prove that $$2(a^3+b^3+c^3)+3\ge 3(ab^2+bc^2+ca^2).$$

However, I only managed to use weighted AM-GM inequality $$ \frac{\frac{2}{3}a^3 + \frac{4}{3}b^3 + abc }{\frac{2}{3}+\frac{4}{3}+1} \ge a^{3/3}b^{5/3}c^{1/3}= \left(ab^2\right)^{2/3} $$ and its cyclic permutations $a\rightarrow b$, $b\rightarrow c$ and $c\rightarrow a$ to obtain $$ 2(a^3+b^3+c^3)+3\ge 3\left[ \left(ab^2\right)^{2/3}+ \left(bc^2\right)^{2/3}+\left(ca^2\right)^{2/3} \right]. $$ But I do not know how to prove the question without the extra exponent $2/3$. Any hints are appreciated.

Zack Fisher
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    Hint: Use Schur's inequality to deal with $a^3 + b^3 + c^3 + 3abc$. The rest should be obvious. $\quad$ (After we replace 3 with $3abc$, we can drop the condition that $abc=1$) AM-GM alone is doomed to failure, because of the $abc$ term leading to $a^x b^y c^z$, which doesn't exist on the RHS (IE You complain about exponents being different) – Calvin Lin Nov 26 '24 at 17:59
  • @CalvinLin Thanks a lot for the nice solution and the explanation! – Zack Fisher Nov 26 '24 at 18:04
  • Can you write it up as a solution? – Calvin Lin Nov 26 '24 at 18:05
  • @CalvinLin Yes, will try. – Zack Fisher Nov 26 '24 at 18:07

1 Answers1

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The following is based on the hint from Calvin Lin's comment.

Replace the $3$ on the LHS by $3abc$ and apply the following special case of Schur's inequality $$ a^3+b^3+c^3 +3abc \ge (ab^2+bc^2+ca^2)+(a^2b+b^2c+c^2a).\tag{1}\label{1} $$ Then it remains to show $$ (a^3+b^3+c^3) + (a^2b+b^2c+c^2a) \ge 2(ab^2 +bc^2 +ca^2) \tag{2}\label{2} $$ in order to get the wanted inequality by adding $\eqref{1}$ and $\eqref{2}$.

But $$\begin{align} b^3 + a^2b &\ge 2\sqrt{a^2b^4}=2ab^2,\\ c^3 + b^2c &\ge 2\sqrt{b^2c^4}=2bc^2,\\ a^3 + c^2a &\ge 2\sqrt{c^2a^4}=2ca^2.\\ \end{align} $$ Adding the above three gives inequality $\eqref{2}$.

Zack Fisher
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