How prove this inequality $a^2+b^2+c^2+8(ab+bc+ac)+3-10(a+b+c)\ge 0$ let $a,b,c\ge 0$,and such $abc=1$,show that

$$a^2+b^2+c^2+8(ab+bc+ac)+3-10(a+b+c)\ge 0$$

My solution: Without loss of generality,assume that

$a=\max{(a,b,c)}$, since $abc=1$,we have
  $a\ge 1$,
we will show that
  $$f(a,b,c)\ge f(a,t,t)\ge 0, t=\sqrt{bc},0<t\le 1$$
since $$f(a,b,c)-f(a,t,t)=(\sqrt{b}-\sqrt{c})^2[(\sqrt{b}+\sqrt{c})^2+8a-10]$$
  then equivalent to
  $$(\sqrt{b}+\sqrt{c})^2+8a\ge 10$$
  which is true because 
  $$(\sqrt{b}+\sqrt{c})^2+8a\ge 4\sqrt{bc}+8a=4(a+\sqrt{bc})+4a\ge 8\sqrt{a\sqrt{bc}}+4a=8\sqrt[4]{a}+4a\ge 12$$

Now,since $a=\dfrac{1}{t^2}$,we have
$$f(a,t,t)=f(\dfrac{1}{t^2},t,t)=\dfrac{(10t^4-7t^2+2t+1)(t-1)^2}{t^4}$$
which is clearly nonnegative because

$$10t^4-7t^2+2t+1=(3t^2-1)^2+t(1-t)+t^4+t>0$$

Have other nice methods? Thank you 

Let $f(x) = x^2+\dfrac8x + 1 - 10x + 16 \log x$ for $x > 0$.
Then the given inequality is $f(a)+f(b)+f(c) \ge 0$, and it is sufficient to show $f(x) \ge 0$.
We note $f'(x) = \dfrac{2(x-2)^2 (x-1)}{x^2}$.
Thus for $x < 1, f'(x)<0$ and for $x > 1, f'(x) \ge 0$.  Hence $\forall x >0, \; f(x) \ge f(1) = 0$.
