Từ gt => \(\hept{\begin{cases}\left(\frac{1}{\sqrt{2}}-\sqrt{x}\right)\left(\frac{1}{\sqrt{2}}-\sqrt{y}\right)\ge0\Leftrightarrow\sqrt{x}+\sqrt{y}\le\frac{\sqrt{2}}{2}+\sqrt{2}\sqrt{xy}\left(1\right)\\x\sqrt{x}\le x\cdot\frac{1}{\sqrt{2}};y\sqrt{y}\le y\cdot\frac{1}{\sqrt{2}}\Rightarrow x\sqrt{x}+y\sqrt{y}\le\frac{1}{\sqrt{2}}\left(x+y\right)\left(2\right)\end{cases}}\)

Lại có \(\hept{\begin{cases}\sqrt{xy}\le xy+\frac{1}{4}\\\sqrt{xy}\le\frac{x+y}{2}\end{cases}\Rightarrow\hept{\begin{cases}\frac{2\sqrt{2}}{3}\sqrt{xy}\le\frac{2\sqrt{2}}{3}\left(xy+\frac{1}{4}\right)\left(3\right)\\\frac{\sqrt{2}}{3}\sqrt{xy}\le\frac{\sqrt{2}}{6}\left(x+y\right)\left(4\right)\end{cases}}}\)

Từ (1)(2)(3)(4) ta có:\(x\sqrt{x}+y\sqrt{y}+\sqrt{x}+\sqrt{y}\le\frac{\sqrt{2}}{2}\left(x+y\right)+\frac{\sqrt{2}}{2}+\frac{2\sqrt{2}}{3}\left(xy+\frac{1}{4}\right)+\frac{\sqrt{2}}{6}\left(x+y\right)\)

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

=> \(VT=\frac{\sqrt{x}}{1+y}+\frac{\sqrt{y}}{1+x}=\frac{x\sqrt{x}+y\sqrt{y}+\sqrt{x}+\sqrt{y}}{1+x+y+xy}\le\frac{2\sqrt{2}}{3}\)

Dấu "=" xảy ra <=> x=y=\(\frac{1}{2}\)

Từ gt => \(\hept{\begin{cases}\left(\frac{1}{\sqrt{2}}-x\right)\left(\frac{1}{\sqrt{2}}-y\right)\ge0\Leftrightarrow\sqrt{x}+\sqrt{y}\le\frac{\sqrt{2}}{2}+\sqrt{2}\sqrt{xy}\left(1\right)\\x\sqrt{x}\le x\cdot\frac{1}{\sqrt{2}};y\sqrt{y}\le y\cdot\frac{1}{\sqrt{2}}\Rightarrow x\sqrt{x}+y\sqrt{y}\le\frac{1}{\sqrt{2}}\left(x+y\right)\left(2\right)\end{cases}}\)

Lại có \(\hept{\begin{cases}\sqrt{xy}\le xy+\frac{1}{4}\\\sqrt{xy}\le\frac{x+y}{2}\end{cases}\Rightarrow\hept{\begin{cases}\frac{2\sqrt{2}}{3}\sqrt{xy}\le\frac{2\sqrt{2}}{3}\left(xy+\frac{1}{4}\right)\left(3\right)\\\frac{\sqrt{2}}{3}\sqrt{xy}\le\frac{\sqrt{2}}{6}\left(x+y\right)\left(4\right)\end{cases}}}\)

Từ (1)(2)(3) và (4) ta có:

\(x\sqrt{x}+y\sqrt{y}+\sqrt{x}+\sqrt{y}\le\frac{\sqrt{2}}{2}\left(x+y\right)+\frac{\sqrt{2}}{2}+\frac{2\sqrt{2}}{3}\left(xy+\frac{1}{4}\right)+\frac{\sqrt{2}}{6}\left(x+y\right)\)

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

=> \(VT=\frac{\sqrt{x}}{1+y}+\frac{\sqrt{y}}{1+x}=\frac{x\sqrt{x}+y\sqrt{y}+\sqrt{x}+\sqrt{y}}{1+x+y+xy}\le\frac{2\sqrt{2}}{3}\)

Dấu "=" xảy ra <=> \(x=y=\frac{1}{2}\)

a/ \(\frac{1}{1+x}+\frac{1}{1+y}\le\frac{2}{1+\sqrt{xy}}\)

\(\Leftrightarrow\left(1+x\right)\left(1+\sqrt{xy}\right)+\left(1+y\right)\left(1+\sqrt{xy}\right)-2\left(1+x\right)\left(1+y\right)\le0\)

\(\Leftrightarrow x\sqrt{xy}+2\sqrt{xy}+y\sqrt{xy}-x-y-2xy\le0\)

\(\Leftrightarrow\sqrt{xy}\left(x-2\sqrt{xy}+y\right)-\left(x-2\sqrt{xy}+y\right)\le0\)

\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{xy}-1\right)\le0\) đúng vì \(x,y\le1\)

b/ Vì \(\hept{\begin{cases}0\le x\le y\le z\le t\\yt\le1\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}xz\le1\\yt\le1\end{cases}}\)

Áp dụng câu a ta được

\(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}+\frac{1}{1+t}\le\frac{2}{1+\sqrt{xz}}+\frac{2}{1+\sqrt{yt}}\le\frac{4}{1+\sqrt[4]{xyzt}}\)

khó quá

KHÓ CHỨ DỄ ĐĂNG LÀM CHI

nhân thêm.

\(\frac{1}{\sqrt{1+x^2}}=\frac{\sqrt{yz}}{\sqrt{yz+x.xyz}}=\sqrt{\frac{yz}{yz+x\left(x+y+z\right)}}=\sqrt{\frac{yz}{\left(x+y\right)\left(x+z\right)}}\)( quen thuộc chưa :v)

CÔSI ta có VT<=1/xy+1/zy+1/zx. 

sau đó vẫn áp dụng bất đẳng thức cosi tùng đôi một vế phải đã cho ta sẽ đc điều phải chứng minh