Power Mean Inequality based question For $a,b,c \in R^+$ prove that  $8(a^3 + b^3 + c^3)^2 \geq 9(a^2 + bc)(b^2 + ca)(c^2 + ab)$.
Well this one is supposed to be done using power-mean inequality, Weighted means and regular means.
Soo power mean tells us that $(\frac{a^3 + b^3 + c^3}{3})^2 \geq (\frac{a^2+b^2+c^2}{3})^3$.
now we have 3 factors of $a^2+b^2+c^2$ so it would be very very nice if we could show that each of these   $\geq$ the three factors on the RHS. however we get it as $a^2+b^2+c^2 \geq a^2 +2bc$ and not $a^2+bc$ making my life very much harder. Now i think taking that extra factor from each mean (namely $bc,ca,ab$) and multiplying them out and separating them from the desired inequality and expressing them as squares which is $\geq 0$ does seem like a way. However even if it is correct its too much brute force and m sure there exists something smarter about which i would love to know. It would be fantastic to see a solution of your own as well as continuation of my work (which will help me learn to clean my mess ;). Thanks a lot.
By power mean inequality,
$\begin{align} (\frac{a^3 + b^3 + c^3}{3})^2 &amp;\geq (\frac{a^2+b^2+c^2}{3})^3 \text{ Or} \\  
8\cdot(\frac{a^3 + b^3 + c^3}{3})^2 &amp; \geq (\frac{2a^2+2b^2+2c^2}{3})^3   \end{align}$
But $2a^2+2b^2+2c^2 \geq a^2+b^2+c^2+ab+bc+ca$ as $a^2+b^2+c^2 \geq ab+bc+ca$. Thus,
$\begin{align} \Biggl( 8\cdot(\frac{a^3 + b^3 + c^3}{3})\Biggr)^2 &amp;\geq \Biggl({\frac{(a^2+bc)+(b^2+ca)+(c^2+ab)}{3}} \Biggr) ^3 \geq \Biggl(\sqrt[3]{(a^2+bc)(b^2+ca)(c^2+ab)}\Biggr) ^{3}=(a^2+bc)(b^2+ca)(c^2+ab) \\
\therefore 8\cdot(a^3 + b^3 + c^3)^2 &amp;\geq 9(a^2+bc)(b^2+ca)(c^2+ab) \\\end{align} $
And we are done.
\begin{eqnarray*}
(8a^2+7bc)(a^2-bc)^2+(8b^2+7ca)(b^2-ca)^2 +(8c^2+7ab)(c^2-ab)^2 \geq 0
\end{eqnarray*}
and rearrange to
\begin{eqnarray*}
8(a^6+b^6+c^6 +2(a^3b^3+b^3c^3+c^3a^3)) \geq 9(a^3b^3+b^3c^3+c^3a^3+a^4bc+b^4ca+c^4ab+2a^2b^2c^2).
\end{eqnarray*}