ta có $x\sqrt{8y-5}\leq \sqrt{3} x\sqrt{\frac{8y-5}{3}} \leq \frac{\sqrt{3}}{2}x(\frac{8y-5}{3}+1)=\frac{\sqrt{3}}{3}x(4y-1)$ $\Rightarrow VT(1) \leq \frac{\sqrt{3}}{3}(8xy-x-y) \leq \frac{\sqrt{3}}{3}(2(x+y)^{2}-x-y)$
VP $\geq \sqrt[4]{24(\frac{1}{2}((x+y)^{2}+4))} =\sqrt[4]{12(x+y)^{2}+96}$
ta cần cm $\sqrt[4]{12(x+y)^{2}+96}\geq \frac{\sqrt{3}}{3} (2(x+y)^{2}-x-y)$
đặt $t=x+y$ $\Leftrightarrow \sqrt[4]{12t^{2}+96}\geq \frac{\sqrt{3}}{3}(2t^{2}-1)$ nâng lũy thừa bậc 4
$\Leftrightarrow (t-2) ( t^{7}+\frac{3t}{2}^{5}+\frac{5t}{2}^{4}+\frac{8t}{16}^{3}+\frac{81t}{8}^{2}+\frac{27t}{2}+2)\leq 0$
$\Leftrightarrow t\leq 2$
ta cần tìm đk này từ pt (2) vs t-x=y
(2) $\Leftrightarrow 11x^{2}-6x(t-x)+3(t-x)^{2}=12x-4(t-x)$
$\Leftrightarrow 20x^{2}-(16+12t)x+3t^{2}+4t=0$
$\Delta' \geq 0 \Leftrightarrow t\in \left[ \frac{-4}{3}{;}2 \right]$
$\Rightarrow x=y=1$