Cho hai số thực dương x, y thỏa mãn

* Hướng dẫn giải

${\log }_{\sqrt{2}}2021=1+{\log }_{\sqrt{2}}\left(\sqrt{1+x^2}+x\right)-{\log }_2\left(y^2-y\sqrt{y^2+2}+1\right)$

$\iff {\log }_2{2021}^2+{\log }_2\left(y^2-y\sqrt{y^2+2}+1\right)={\log }_{2}2+{\log }_2{\left(\sqrt{1+x^2}+x\right)}^2$

$\iff {2021}^2\left(y^2-y\sqrt{y^2+2}+1\right)=2{\left(\sqrt{1+x^2}+x\right)}^2$

$\iff {2021}^2\left(2y^2-2y\sqrt{y^2+2}+2\right)=4{\left(\sqrt{1+x^2}+x\right)}^2$

$\iff {2021}^2{\left(\sqrt{y^2+2}-y\right)}^2=4{\left(\sqrt{1+x^2}+x\right)}^2$

$\iff 2021\left(\sqrt{y^2+2}-y\right)=2\left(\sqrt{1+x^2}+x\right)\Rightarrow \left\{\begin{array}{l}\left(\sqrt{1+x^2}+x\right)\left(\sqrt{1+y^2}+y\right)=2021\left(1\right)\\ \left(\sqrt{1+x^2}-x\right)\left(\sqrt{y^2+2}-y\right)=\frac{2}{2021}\left(2\right)\end{array}\right.$

Cộng 2 vế (1) và (2) ta được:

$2021+\frac{2}{2021}=2\sqrt{\left(1+x^2\right)\left(y^2+2\right)}+2xy\le 1+x^2+y^2+2+2xy={\left(x+y\right)}^2+3$

$\Rightarrow x+y\ge \sqrt{2021+\frac{2}{2021}-3}\approx 44,922.$

Chọn C.