[hongduc_cqt]

Mình có 1 cách khác:
Cộng vế với vế ta có:
$\sqrt{(\sin x)^2+\dfrac{1}{\sin x)^2}}+\sqrt{(\cos x)^2+\dfrac{1}{\cos x)^2}}+\sqrt{(\sin y)^2+\dfrac{1}{\sin y)^2}}+\sqrt{(\cos y)^2+\dfrac{1}{\cos y)^2}} = \sqrt{20}.\sqrt{\dfrac{x}{x+y}}+\sqrt{\dfrac{y}{x+ y}}
Ta có $VP^2= \sqrt{20}.(1+\sqrt{\dfrac{xy}{(x+y)^2}}\leq 40
\Rightarrow VT \leq 2\sqrt{10}$
Ta cm $VT \geq 2\sqrt(10)$
$VT= \sqrt{(\sin x)^2+\dfrac{1}{\sin x)^2}}+\sqrt{(\cos x)^2+\dfrac{1}{\cos x)^2}}$
$= 1+\dfrac{4}{\sin(2x)}+2.\sqrt{\dfrac{\sin{2x}}{4}+ (\tan x)^2+\dfrac{1}{(\tan x)^2}+\dfrac{4}{\sin(2x)}}\geq 2\sqrt(10)$ ($\sin(2x)\leq 1$)
Ta có dấu $=$ xảy ra khi: $x=y=\dfrac{\pi}{4} +k2\pi$.
[RIGHT][I][B]Nguồn: MathScope.ORG[/B][/I][/RIGHT]