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.
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.
I remember I solved this inequality a few years ago. Here is my proof. It may look complicated but the idea is actually very simple - we want to make $b,c$ closer to each other keeping the inequality $a^2\ge 36bc$ valid and the sum $b^2+c^2$ fixed so the lefthandside remains fixed and the righthandside increases. The process of making $b,c$ closer to each other will terminate either in the moment when $36bc$ becomes equal to $a^2$ (this corresponds to the second case below) or when $b=c$ (this is the first case).
Let $d=\sqrt{\frac{b^2+c^2}2}$ so that $2d^2=b^2+c^2$. Observe that $d^2 = \frac{b^2+c^2}2 \ge bc$ and $2d = \sqrt{2(b^2+c^2)}\ge b+c$. Thus $$d^2+2ad \ge bc+a(b+c)=ab+bc+ca.$$ It is enough to prove that $$\frac{a^2}3+2d^2 > d^2+2ad.$$ This is true if $a\ge 6d$ because $$\frac{a^2}3+2d^2 - (d^2+2ad) = \frac{a(a-6d)}{3}+d^2>0.$$
We are left with the case $a<6d$.
Due to $a<6d$ there are $m,n>0$ such that $m^2+n^2=2d^2$ and $36mn=a^2$. Then $36bc<a^2=36mn$ so $bc<mn$ and also $$b+c=\sqrt{b^2+2bc+c^2}=\sqrt{2d^2+2bc}<\sqrt{2d^2+2mn}=\sqrt{m^2+2mn+n^2}=m+n.$$ It follows that $$ab+bc+ca=a(b+c)+bc<a(m+n)+mn=6(m+n)\sqrt{mn}+mn$$ so we only have to prove that $$\frac{a^2}3+b^2+c^2=12mn + n^2+m^2 \ge 6(m+n)\sqrt{mn} + mn,$$ which is true because $$12mn + n^2+m^2 - (6(m+n)\sqrt{mn} + mn) = (m+n-3\sqrt{mn})^2\ge 0.$$
