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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 tính chất của phân số, ta có:

\(\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}< \frac{\sqrt{x}+\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\) ; \(\frac{\sqrt{y}}{\sqrt{y}+\sqrt{z}}< \frac{\sqrt{y}+\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\); \(\frac{\sqrt{z}}{\sqrt{z}+\sqrt{x}}< \frac{\sqrt{z}+\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Cộng vế với vế:

\(\Rightarrow VT< \frac{2\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=2\) (đpcm)

ĐỊT MẸ

Ta có: \(\frac{1}{2}.2x\left(1-x\right)\left(1-x\right)\le\frac{1}{2}\left[\frac{2x+1-x+1-x}{3}\right]^3=\frac{4}{27}\)

\(\Rightarrow\sqrt{x}\left(1-x\right)\le\frac{2\sqrt{3}}{9}\Rightarrow\frac{1}{\sqrt{x}\left(1-x\right)}\ge\frac{9}{2\sqrt{3}}\)

\(\Rightarrow\frac{\sqrt{x}}{1-x}\ge\frac{3\sqrt{3}}{2}x\). Thiết lập tương tự hai BĐT còn lại và cộng theo vế thu được đpcm.

Áp dụng BĐT Cô si ta có:

\(x+y\ge2\sqrt{xy}=2\cdot\frac{1}{\sqrt{z}};y+z\ge2\sqrt{yz}=2\cdot\frac{1}{\sqrt{x}};z+x\ge2\sqrt{xz}=2\cdot\frac{1}{\sqrt{y}}.\)( vì xyz=1)

=> P\(\ge\)\(\frac{2x\sqrt{x}}{y\sqrt{y}+2z\sqrt{z}}\)+ \(\frac{2y\sqrt{y}}{z\sqrt{z}+2x\sqrt{x}}+\frac{2z\sqrt{z}}{x\sqrt{x}+2y\sqrt{y}}\)

Đặt \(\hept{\begin{cases}a=y\sqrt{y}+2z\sqrt{z}\\b=z\sqrt{z}+2x\sqrt{x}\\c=x\sqrt{x}+2y\sqrt{y}\end{cases}\left(a;b;c\ge0\right)}\)<=> \(\hept{\begin{cases}4a+b=2c+9z\sqrt{z}\\4b+c=2a+9x\sqrt{x}\\4c+a=2b+9y\sqrt{y}\end{cases}}\)

<=> \(\hept{\begin{cases}z\sqrt{z}=\frac{4a+b-2c}{9}\\x\sqrt{x}=\frac{4b+c-2a}{9}\\y\sqrt{y}=\frac{4c+a-2b}{9}\end{cases}}\)

Do đó:

P \(\ge\)\(\frac{2}{9}\cdot\left(\frac{4a+b-2c}{c}+\frac{4b+c-2a}{a}+\frac{4c+a-2b}{b}\right)\)

<=> P \(\ge\)\(\frac{2}{9}\left(4\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)+\left(\frac{b}{c}+\frac{c}{a}+\frac{a}{b}\right)-6\right)\)

<=> P \(\ge\frac{2}{9}\cdot\left(4\cdot3\cdot\sqrt[3]{\frac{a}{c}\cdot\frac{b}{a}\cdot\frac{c}{b}}+3\cdot\sqrt[3]{\frac{b}{c}\cdot\frac{c}{a}\cdot\frac{a}{b}}-6\right)\)( Áp dụng BĐT Cô si cho 3 số ko âm)

<=> P \(\ge\frac{2}{9}\left(12+3-6\right)=2\)( đpcm)

Dấu = khi x=y=z=1.

Đặt \(\sqrt{x}=a\) , \(\sqrt{y}=b\) , \(\sqrt{z}=c\)

Suy ra \(P=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-c\right)\left(b-a\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)

\(=-\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

Xét tử : \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)=a^2\left[-\left(a-b\right)-\left(c-a\right)\right]+b^2\left(c-a\right)+c^2\left(a-b\right)\)

\(=\left(a-b\right)\left(c^2-a^2\right)+\left(c-a\right)\left(b^2-a^2\right)=\left(a-b\right)\left(c-a\right)\left(c+a\right)+\left(c-a\right)\left(b-a\right)\left(b+a\right)\)

\(=\left(a-b\right)\left(c-a\right)\left(c+a-a-b\right)=\left(a-b\right)\left(c-a\right)\left(c-b\right)\)

Suy ra \(P=-\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)