<s>[INST] How can I prove this inequality: $\frac{a^2}{3}+b^2+c^2>ab+bc+ca$? If $abc=1$ and $a^3>36$ prove that,
$\frac{a^2}{3}+b^2+c^2>ab+bc+ca$

I tried to use the general proof method.
$\frac{a^2}{3}+b^2+c^2-ab-bc-ca>0$

Symbolically notation:
$\frac{a^2}{3}+b^2+c^2-ab-bc-ca>0 \Rightarrow(x+y+z+...)^2>0$
But, after trying too much, I accepted the defeat.
 [/INST] As an old (not too old) Olympiad participant, I love the questions of Proof. My reputation is very low right now.But, I think the proof is correct.
\begin{align}\frac{a^2}{3}+b^2+c^2-ab-bc-ca&=\frac{a^2}{4}+b^2+c^2-ab-bc-ca+\frac{a^2}{12}\\
&=\left({\frac a2-b-c}\right)^2+\frac{a^2}{12}-3bc\\
&=\left({\frac a2-b-c}\right)^2+\frac{a^2}{12}-\frac 3a\\
&=\left({\frac a2-b-c}\right)^2+\frac{a^3-36}{12a}\\
&>0
\end{align}
Done.
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