Theo giả thiết ta có \(\sqrt{x}+\sqrt{y}-\sqrt{z}=0\)

                            \(\Leftrightarrow\sqrt{x}+\sqrt{y}=\sqrt{z}\)

                            \(\Leftrightarrow x+y+2\sqrt{xy}=z\)

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

         Tương tự \(x+z-y=2\sqrt{xz}\)     ;    \(y+z-x=2\sqrt{yz}\)

    Suy ra  \(\frac{1}{x+y-z}+\frac{1}{x+z-y}+\frac{1}{y+z-x}=\frac{1}{-2\sqrt{xy}}+\frac{1}{2\sqrt{xz}}+\frac{1}{2\sqrt{yz}}=\frac{\sqrt{x}+\sqrt{y}-\sqrt{z}}{2\sqrt{xyz}}=0\)

   Vậy suy ra ĐPCM , bạn ghi nhầm đề đúng ko

@Trần Huỳnh Thanh Long sai đề ở đâu ạ

\(P=\frac{x\left(\sqrt{y}-\sqrt{z}\right)-y\left(\sqrt{x}-\sqrt{z}\right)+z\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{z}\right)}\)

\(P=\frac{x\left(\sqrt{y}-\sqrt{z}\right)-y\left[\left(\sqrt{y}-\sqrt{z}\right)+\left(\sqrt{x}-\sqrt{y}\right)\right]+z\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{z}\right)}\)

\(P=\frac{\left(x-y\right)\left(\sqrt{y}-\sqrt{z}\right)+\left(z-y\right)\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{z}\right)}\)

\(P=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)-\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{z}\right)}\)

\(P=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left[\left(\sqrt{x}+\sqrt{y}\right)-\left(\sqrt{y}+\sqrt{z}\right)\right]}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{z}\right)}\)

\(P=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{z}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{z}\right)}=1\)

=> đpcm

ĐỊT MẸ

Sử dụng BĐT AM-GM, ta có: 

\(x^3+y^2\ge2yx\sqrt{x}\)

\(\Rightarrow\frac{2\sqrt{x}}{x^3+y^2}\le\frac{2\sqrt{x}}{2yx\sqrt{x}}=\frac{1}{xy}\)

Tương tự cộng lại suy ra: 

\(VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)

Để ý: \(2=\left(\frac{x}{y}+\frac{y}{z}\right)+\left(\frac{y}{z}+\frac{z}{x}\right)+\left(\frac{z}{x}+\frac{x}{y}\right)\)

\(\ge2\sqrt{\frac{x}{z}}+2\sqrt{\frac{y}{x}}+2\sqrt{\frac{z}{y}}\)

Từ đó suy ra \(\sqrt{\frac{y}{x}}+\sqrt{\frac{z}{y}}+\sqrt{\frac{x}{z}}\le1\)

ĐKXĐ: ...

Lấy pt cuối trừ 3 lần pt đầu ta được:

\(\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^3+\left(\sqrt{y}-\frac{1}{\sqrt{y}}\right)^3+\left(\sqrt{z}-\frac{1}{\sqrt{z}}\right)^3=\frac{512}{27}\)

Pt (2) tương đương:

\(x+\frac{1}{x}-2+y+\frac{1}{y}-2+z+\frac{1}{z}-2=\frac{64}{9}\)

\(\Leftrightarrow\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2+\left(\sqrt{y}-\frac{1}{\sqrt{y}}\right)^2+\left(\sqrt{z}-\frac{1}{\sqrt{z}}\right)^2=\frac{64}{9}\)

Đặt \(\left(\sqrt{x}-\frac{1}{\sqrt{x}};\sqrt{y}-\frac{1}{\sqrt{y}};\sqrt{z}-\frac{1}{\sqrt{z}}\right)=\left(a;b;c\right)\)

Hệ trở thành:

\(\left\{{}\begin{matrix}a+b+c=\frac{8}{3}\\a^2+b^2+c^2=\frac{64}{9}\\a^3+b^3+c^3=\frac{512}{27}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b+c=\frac{8}{3}\\ab+bc+ca=0\\a^3+b^3+c^3=\frac{512}{27}\end{matrix}\right.\)

Ta có: \(a^3+b^3+c^3-3abc=\frac{512}{27}-3abc\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=\frac{512}{27}-3abc\)

\(\Leftrightarrow\frac{8}{3}.\left(\frac{64}{9}-0\right)=\frac{512}{27}-3abc\)

\(\Rightarrow abc=0\)

\(\Rightarrow\left\{{}\begin{matrix}a+b+c=\frac{8}{3}\\ab+bc+ca=0\\abc=0\end{matrix}\right.\) \(\Leftrightarrow\left(a;b;c\right)=\left(0;0;\frac{8}{3}\right)\) và hoán vị

Hay \(\left(x;y;z\right)=\left(1;1;9\right)\) và hoán vị

Theo AM-GM: \(x^3+y^2\ge2\sqrt{x^3y^2}=2xy\sqrt{x}\)

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

Tương tự: \(\frac{2\sqrt{y}}{y^3+z^2}\le\frac{1}{yz}\)

\(\frac{2\sqrt{z}}{z^3+x^2}\le\frac{1}{zx}\)

Cộng vế với vế => \(VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\)

Theo AM-GM; \(VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\le\frac{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{z^2}+\frac{1}{x^2}}{2}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)

Dấu " = " xảy ra <=> x=y=z=1

Áp dụng bất đẳng thức Cacuhy - Schwarz 

\(\Rightarrow\hept{\begin{cases}x^3+y^2\ge2\sqrt{x^3y^2}=2xy\sqrt{x}\\y^3+z^2\ge2\sqrt{y^3z^2}=2yz\sqrt{y}\\z^3+x^2\ge2\sqrt{z^3x^2}=2xz\sqrt{z}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\frac{2\sqrt{x}}{x^3+y^2}\le\frac{2\sqrt{x}}{2xy\sqrt{x}}=\frac{1}{xy}\\\frac{2\sqrt{y}}{y^3+z^2}\le\frac{2\sqrt{y}}{2yz\sqrt{y}}=\frac{1}{yz}\\\frac{2\sqrt{z}}{z^3+x^2}\le\frac{2\sqrt{z}}{2xz\sqrt{z}}=\frac{1}{xz}\end{cases}}\)

\(\Rightarrow VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\left(1\right)\)

Áp dụng bất đẳng thức Cacuchy Schwarz 

\(\Rightarrow\hept{\begin{cases}\frac{1}{x^2}+\frac{1}{y^2}\ge2\sqrt{\frac{1}{x^2y^2}}=\frac{2}{xy}\\\frac{1}{y^2}+\frac{1}{z^2}\ge2\sqrt{\frac{1}{y^2z^2}}=\frac{2}{yz}\\\frac{1}{z^2}+\frac{1}{x^2}\ge2\sqrt{\frac{1}{z^2x^2}}=\frac{2}{xz}\end{cases}}\)

\(\Rightarrow2\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\ge2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\)

\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\left(2\right)\)

Từ (1) và (2)

\(\Rightarrow VT\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)

\(\Leftrightarrow\frac{2\sqrt{x}}{x^3+y^2}+\frac{2\sqrt{y}}{y^3+z^2}+\frac{2\sqrt{z}}{z^3+x^2}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\left(đpcm\right)\)