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 

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.
