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 
 A: Here is a straightforward (but not very elegant) solution : let us put
$$
D=3(a^4+b^4+c^4)+2(a+b+c)abc - 5(a^2b^2+b^2c^2+a^2c^2)
$$
We may assume without loss that $a\leq b\leq c$.
Then, if we put $u=b-a$ and $v=c-b$, we have
$$
D=u^4 + (8a + 2v)u^3 + (10a^2 + 12va + 13v^2)u^2 + (10va^2 + 28v^2a + 12v^3)u + (10v^2a^2 + 12v^3a + 3v^4)
$$
and we are done since all the coefficients are positive.
A: 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.
