Find minimum of $P=\frac{\sqrt{3(2x^2+2x+1)}}{3}+\frac{1}{\sqrt{2x^2+(3-\sqrt{3})x +3}}+\frac{1}{\sqrt{2x^2+(3+\sqrt{3})x +3}}$ For $x\in\mathbb{R}$ find minimum of $P$.
$P=\dfrac{\sqrt{3(2x^2+2x+1)}}{3}+\dfrac{1}{\sqrt{2x^2+(3-\sqrt{3})x +3}}+\dfrac{1}{\sqrt{2x^2+(3+\sqrt{3})x +3}}$
Source : Viet Nam national test for high school student. 
I can't slove this problem.
 A: note when $x=0$, the three items are equal which hints the min is $\sqrt{3}$.
first take $\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}} \ge \sqrt{\dfrac{8}{a+b}}$ which is easy to prove by AM-GM.
$P \ge \dfrac{\sqrt{3(2x^2+2x+1)}}{3}+\sqrt{\dfrac{8}{4x^2+6x+6}} \ge \dfrac{\sqrt{3(2x^2+2x+1)}}{3}+\sqrt{\dfrac{8}{6x^2+6x+6}} $, when $x=0$ get "="
let $t=\sqrt{x^2+x+1},t\ge \dfrac{\sqrt{3}}{2} \implies P\ge \sqrt{\dfrac{2}{3}}\left( \sqrt{t^2-\dfrac{1}{2}}+\dfrac{\sqrt{2}}{t} \right)$
now we prove $ \sqrt{t^2-\dfrac{1}{2}}+\dfrac{\sqrt{2}}{t} \ge \dfrac{3}{\sqrt{2}} \iff \sqrt{t^2-\dfrac{1}{2}}\ge \dfrac{3}{\sqrt{2}}-\dfrac{\sqrt{2}}{t} >0 \iff t^2-\dfrac{1}{2} \ge \dfrac{9}{2}-\dfrac{6}{t}+\dfrac{2}{t^2} \iff t^2(t^2-1) \ge 4t^2-6t+2 \iff (t-1)(t^2(t+1)-2(2t-1)) \ge 0 \iff (t-1)(t(t^2+t-2)-2(t-1))\iff (t-1)^2(t^2+2t-2) \ge 0 \iff (t-1)^2(t+1+\sqrt{3})(t+1-\sqrt{3})\ge 0$
it is trivial $(t+1+\sqrt{3})>0,(t+1-\sqrt{3}) \ge \dfrac{\sqrt{3}}{2}+1-\sqrt{3} =1-\dfrac{\sqrt{3}}{2}>0 $
so when $t=1, P_{min}= \sqrt{3}$
$t=1 \implies x=0 or x=-1$,since first two "=" are all take $x=0$ which mean $P$ get min when $x=0$.
BTW, it may be hard to prove a stronger one:
$P \ge \sqrt{\dfrac{x^2}{3}+3}$
