Từ \(0\le x\le y\le1\) và \(2x+y\le2\Rightarrow2x^2+xy\le2x\)(nhân cả 2 vế với \(x\ge0\))

                                                                  \(\left(y-x\right)y\le y-x\)(nhân cả 2 vế của \(0\le y\le1\)với \(y-x\ge0\)(do \(x\le y\))

Cộng từng vế ta có : 

\(2x^2+xy+\left(y-x\right)y\le2x+y-x\)

\(\Leftrightarrow2x^2+y^2\le x+y\)

\(\Leftrightarrow\left(2x^2+y^2\right)^2\le\left(x+y\right)^2\)

Mặt khác \(\left(x+y\right)^2=\left(\frac{1}{\sqrt{2}}.\sqrt{2}x+1.y\right)^2\le\left(\frac{1}{2}+1\right)\left(2x^2+y^2\right)\)(bất đẳng thức Bunhiacopxki)

\(\Rightarrow\left(2x^2+y^2\right)^2\le\frac{3}{2}\left(2x^2+y^2\right).\)

\(\Leftrightarrow2x^2+y^2\le\frac{3}{2}.\)(đpcm)

Chúc học tốt 

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 <=> \(x=y=\frac{1}{2}\)

Ta có \(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+xz\right)=4\)

=> \(\orbr{\begin{cases}x+y+z=2\\x+y+z=-2\end{cases}}\)

+ \(x+y+z=2\)

Thay vào Pt (1)

=> \(xy+z\left(2-z\right)=1\)

 => \(xy=\left(z-1\right)^2\)=> \(x,y,z\ge0\)( do \(x+y+z=2>0\))

Mà \(xy\le\left(\frac{x+y}{2}\right)^2=\left(\frac{2-z}{2}\right)^2\)

=> \(z-1\le\frac{2-z}{2}\)=> \(z\le\frac{4}{3}\)

Hoàn toàn TT => \(x,y,z\le\frac{4}{3}\)

+ \(x+y+z=-2\)

=> \(xy+z\left(-2-z\right)=1\)

=> \(xy=\left(z+1\right)^2\)=> \(x,y,z\le0\)( do \(x+y+z=-2< 0\))

Mà \(xy\le\left(\frac{x+y}{2}\right)^2=\left(\frac{-2-z}{2}\right)^2\)

=> \(\left(z+1\right)^2\le\left(\frac{z+2}{2}\right)^2\)

=> \(z+1\ge\frac{-z-2}{2}\)=> \(z\ge-\frac{4}{3}\)

TT => \(x,y,z\ge-\frac{4}{3}\)

Vậy \(-\frac{4}{3}\le x,y,z\le\frac{4}{3}\)