Mình nghĩ đề bài là tìm GTLN :D Sai thì thôi :D
\(\dfrac{1}{1+\sqrt{x}}+\dfrac{1}{1+\sqrt{y}}+\dfrac{1}{1+\sqrt{z}}=2\)
⇔ \(\dfrac{1}{1+\sqrt{x}}=1-\dfrac{1}{1+\sqrt{y}}+1-\dfrac{1}{1+\sqrt{z}}=\dfrac{\sqrt{y}}{1+\sqrt{y}}+\dfrac{\sqrt{z}}{1+\sqrt{z}}\text{≥}2\sqrt{\dfrac{\sqrt{yz}}{\left(1+\sqrt{y}\right)\left(1+\sqrt{z}\right)}}\) Làm tương tự : \(\dfrac{1}{1+\sqrt{y}}\text{≥}2\sqrt{\dfrac{\sqrt{xz}}{\left(1+\sqrt{x}\right)\left(1+\sqrt{z}\right)}}\)
\(\dfrac{1}{1+\sqrt{z}}\text{≥}2\sqrt{\dfrac{\sqrt{xy}}{\left(1+\sqrt{x}\right)\left(1+\sqrt{y}\right)}}\)
⇒ \(\dfrac{1}{1+\sqrt{x}}.\dfrac{1}{1+\sqrt{y}}.\dfrac{1}{1+\sqrt{z}}\text{≥}8.\dfrac{\sqrt{xyz}}{\left(1+\sqrt{x}\right)\left(1+\sqrt{y}\right)\left(1+\sqrt{z}\right)}\)
⇔ \(\dfrac{1}{8}\text{≥}\sqrt{xyz}\)
\("="\text{⇔}x=y=z=\dfrac{1}{2}\)
