Gõ lần 2 :(
Theo C-S :
$\frac{1}{\sqrt{2x^{2}+(3-\sqrt{3})x+3}}+\frac{1}{\sqrt{2x^{2}+(3+\sqrt{3})x+3}}$ $\geq \frac{4}{\sqrt{2x^{2}+(3-\sqrt{3})x+3}+ \sqrt{2x^{2}+(3+\sqrt{3})+3}}$
$\geq \frac{4}{\sqrt{2[2x^{2}+(3-\sqrt{3})x+3+2x^{2}+(3+\sqrt{3})x+3]}}$
( A/d $\sqrt{a}+\sqrt{b}\leq \sqrt{2(a+b)}$)
$=\frac{2}{\sqrt{2x^{2}+3x+3}} \geq \frac{2}{\sqrt{3x^{2}+3x+3}}$
=> Có :
$P \geq \frac{\sqrt{3(2x^{2}+2x+1)}}{3}+\frac{2}{\sqrt{3x^{2}+3x+3}}=M$
Ta sẽ CM : $M \geq \sqrt{3}$
$<=> \sqrt{2x^{2}+2x+1}+\frac{2}{\sqrt{x^{2}+x+1}} \geq 3(*)$
Đặt : $3t=\frac{2}{\sqrt{x^{2}+x+1}}$ $\geq 0\Rightarrow $ $2x^{2}+2x+1=\frac{8}{9t^2}-1$
=> Ta cần CM bđt :
$\sqrt{8/(9t^{2})-1} \geq 3-3t$ => luôn đúng với $\forall x\geq 0$
Vậy : $P \geq M \geq \sqrt{3}=> Min P = \sqrt{3}$
Dấu = xảy ra $<=> x=0$