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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) $\Leftrightarrow 11x^{2}-6x(t-x)+3(t-x)=12x-4(t-x)$ $\Leftrightarrow 20x^{2}-(16+12t)x+3t^{2}+4=0$ $\Delta' \geq 0 \Leftrightarrow t\in \left[ \frac{-4}{3}{;}2 \right]$ $\Rightarrow x=y=1$

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) $\Leftrightarrow 11x^{2}-6x(t-x)+3(t-x)^{2}=12x-4(t-x)$ $\Leftrightarrow 20x^{2}-(16+12t)x+3t^{2}+4t=0$ $\Delta' \geq 0 \Leftrightarrow t\in \left[ \frac{-4}{3}{;}2 \right]$ $\Rightarrow x=y=1$

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) $\Leftrightarrow 11x^{2}-6x(t-x)+3(t-x)=12x-4(t-x)$ $\Leftrightarrow 20x^{2}-(16+12y)x+3t^{2}+4=0$ $\Delta' \geq 0 \Leftrightarrow t\in \left[ \frac{-4}{3}{;}2 \right]$ $\Rightarrow x=y=1$

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) $\Leftrightarrow 11x^{2}-6x(t-x)+3(t-x)=12x-4(t-x)$ $\Leftrightarrow 20x^{2}-(16+12t)x+3t^{2}+4=0$ $\Delta' \geq 0 \Leftrightarrow t\in \left[ \frac{-4}{3}{;}2 \right]$ $\Rightarrow x=y=1$

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) $\Leftrightarrow 11x^{2}-6x9t-x)+3(t-x)=12x-4(t-x)$ $\Leftrightarrow 20x^{2}-(16+12y)x+3t^{2}+4=0$ $\Delta' \geq 0 \Leftrightarrow t\in \left[ \frac{-4}{3}{;}2 \right]$ $\Rightarrow x=y=1$

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) $\Leftrightarrow 11x^{2}-6x(t-x)+3(t-x)=12x-4(t-x)$ $\Leftrightarrow 20x^{2}-(16+12y)x+3t^{2}+4=0$ $\Delta' \geq 0 \Leftrightarrow t\in \left[ \frac{-4}{3}{;}2 \right]$ $\Rightarrow x=y=1$