How prove $\sqrt{x}+\sqrt{y}+\sqrt{\frac{x+y+2}{xy-1}}\ge 2\left( \frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\sqrt{\frac{xy-1}{x+y+2}}\right)$ How prove 
$\sqrt{x}+\sqrt{y}+\sqrt{\frac{x+y+2}{xy-1}}\ge 2\left( \frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\sqrt{\frac{xy-1}{x+y+2}}\right)$
for $8x\ge13, 8y \ge 13$?
 A: For easy calculation,let $a=\sqrt{x},b=\sqrt{y} \implies a+b+\sqrt{\dfrac{a^2b^2-1}{a^2+b^2+2}} \ge 2\left(\dfrac{1}{a}+\dfrac{1}{b}+\sqrt{\dfrac{a^2+b^2+2}{a^2b^2-1}}\right) \iff (a+b)\dfrac{ab-2}{ab}+\dfrac{a^2+b^2+4-2a^2b^2}{\sqrt{(a^2b^2-1)(a^2+b^2+2)}} \ge 0$
we have two cases:
1) $ab\ge 2 $
$\dfrac{a^2+b^2+4-2a^2b^2}{\sqrt{(a^2b^2-1)(a^2+b^2+2)}} \ge \dfrac{2ab+4-2a^2b^2}{\sqrt{(a^2b^2-1)(a^2+b^2+2)}}=\dfrac{-2(ab-2)(ab+1)}{\sqrt{(a^2b^2-1)(a^2+b^2+2)}} \iff (ab-2)\left( \dfrac{a+b}{ab}-\dfrac{2(ab+1)}{\sqrt{(a^2b^2-1)(a^2+b^2+2)}}\right) \ge 0  \iff  \\ \dfrac{a+b}{ab}-\dfrac{2(ab+1)}{\sqrt{(a^2b^2-1)(a^2+b^2+2)}} \ge 0 \iff \dfrac{2}{\sqrt{ab}}-\dfrac{2(ab+1)}{\sqrt{(a^2b^2-1)(2ab+2)}}\ge0 \iff \\ \dfrac{1}{\sqrt{ab}} \ge \dfrac{1}{\sqrt{2(ab-1)}} \iff ab\ge2 $
2)$ab < 2$
$\dfrac{a^2+b^2+4-2a^2b^2}{\sqrt{(a^2b^2-1)(a^2+b^2+2)}} \ge (a+b)\dfrac{2-ab}{ab}$
$t=ab<2,p=a^2+b^2 \ge 2t \iff \dfrac{p+4-2t^2}{\sqrt{(t^2-1)(p+2)}}\ge \dfrac{(2-t)\sqrt{p+2}}{t} \iff \\A(t)p^2+B(t)p+C(t) \ge 0 \\ A(t)=(-t^4+4t^3-2t^2-4t+4)=t^2(2-t)+(t-1)^2(2t+2) > 0\\ B(t)=(-2t^5+2t^4+2t^3-6t^2+8) \\ C(t)=4t^6-4t^5-12t^3+16t$
$4tA(t)+B(t)=-6t^5+18t^4-6t^3-22t^2+16t+8  >0 \implies A(t)p^2+B(t)p+C(t) \ge A(t)(2t)^2+B(t)(2t)+C(t)=4t(t+1)^2(2-t)^3 >0$
So the "=" will hold then $a=b, ab=2 \implies x=y=2$
there is no limitation for $x,y$ except $x>0,y>0,xy>1$
QED
