Quay lại đáp án

Đk: $0\leq x;y\leq 1/2$Đặt $u=x\sqrt{2};v=y\sqrt{2};u,v\in \left[ {} \right.0;1/2\left[ {} \right.$$(1)\Leftrightarrow \Sigma \frac{1}{\sqrt{1+u^2}}=\frac{2}{\sqrt{1+uv}},\left\{ \begin{array}{l} u,v\geq 0\\ uv\leq 1\end{array} \right.$ (3)Ta có : $\Sigma 1.\frac{1}{\sqrt{1+u^2}}\leq \sqrt{2}.\sqrt{\frac{1}{1+u^2}+\frac{1}{1+v^2}}$ (4)( Cauchy-Schwarz)Mặt khác: $\forall u,v\in \left[ {} \right.0;1\left[ {} \right.$ thì $\Sigma \frac{1}{1+u^2}\leq \frac{2}{1+uv}$ (5)

Đk: $0\leq x;y\leq 1/2$Đặt $u=x\sqrt{2};v=y\sqrt{2};u,v\in \left[ {} \right.0;1/2\left[ {} \right.$$(1)\Leftrightarrow \Sigma \frac{1}{\sqrt{1+u^2}}=\frac{2}{\sqrt{1+uv}},\left\{ \begin{array}{l} u,v\geq 0\\ uv\leq 1\end{array} \right.$ (3)Ta có : $\Sigma 1.\frac{1}{\sqrt{1+u^2}}\leq \sqrt{2}.\sqrt{\frac{1}{1+u^2}+\frac{1}{1+v^2}}$ (4)( Cauchy-Schwarz)Mặt khác: $\forall u,v\in \left[ {} \right.0;1\left[ {} \right.$ thì $\Sigma \frac{1}{1+u^2}\leq \frac{2}{1+uv}$ (5)Thật vậy: (5) $\Leftrightarrow \Sigma (\frac{1}{1+u^2}-\frac{1}{1+uv})\leq 0$tự c/m nhak!Từ (3); (4); (5) suy ra :$(3)\Leftrightarrow u=v\Rightarrow x=y$$(2)\Leftrightarrow \sqrt{x-2x^2}=1/9\Leftrightarrow x=y=\frac{9+(-)\sqrt{73}}{36}$thỏa mãn đkKL:.........