<s>[INST] How prove $3(a^4+b^4+c^4)+2(a+b+c)abc\ge 5(a^2b^2+b^2c^2+a^2c^2)$ let $a,b,c>0$ and such $abc=1$,show that
$$3(a^4+b^4+c^4)+2(a+b+c)\ge 5\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)$$
my idea: maybe can use AM-GM inequality,
$$3(a^4+b^4+c^4)a^2b^2c^2+2(a+b+c)(a^2b^2c^2)\ge 5(a^2b^2+b^2c^2+a^2c^2)$$
$$\Longleftrightarrow 3(a^4+b^4+c^4)+2(a+b+c)abc\ge 5(a^2b^2+b^2c^2+a^2c^2)$$
then I can't,Thank you very much 
 [/INST] Consider Schur's inequality for $r = 2$:
$$a^4 + b^4 + c^4 + (a + b + c)abc \geqslant a^3b + a^3c + b^3a + b^3c + c^3a + c^3b$$
plus three AM-GMs:
$$
\begin{aligned}
a^3b + b^3a &\geqslant 2a^2b^2\\
b^3c + c^3b &\geqslant 2b^2c^2\\
a^3c + c^3a &\geqslant 2a^2c^2
\end{aligned}
$$
This gives you, after multiplying by $2$:
$$2(a^4 + b^4 + c^4) + 2(a + b + c)abc \geqslant 4(a^2b^2 + b^2c^2 + a^2c^2).$$
Three more AM-GMs:
$$
\begin{align}
\frac{a^4 + b^4}{2} &\geqslant a^2b^2,\\
\frac{b^4 + c^4}{2} &\geqslant b^2c^2,\\
\frac{a^4 + c^4}{2} &\geqslant a^2c^2.
\end{align}$$
Sum up 4 last inequalities and you're done.
Equality, of course, holds only if $a = b = c$, which can be seen from all those AM-GMs.
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