<s>[INST] 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 
 [/INST] Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, our inequality is a linear inequality of $v^2$, which says that
it's remains to prove the last inequality for an extremal value of $v^2$,
which happens for equality case of two variables.
Let $b=a$ and $c=\frac{1}{a^2}$.
Hence, we need to prove that 
$$(a-1)^2(10a^4-7a^2+2a+1)\geq0,$$
which is obvious.
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