dat a=x-y

b=y-z 

c=z-x

a+b+c=0=x+y+z

\(\left(\frac{a}{z}+\frac{b}{x}+\frac{c}{y}\right)\left(\frac{z}{a}+\frac{x}{b}+\frac{y}{c}\right)\)

dung bumiakopsky de giai

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Ta co:\(x+y+z=0\)

\(\Leftrightarrow\frac{x+y+z}{xyz}=0\)

\(\Leftrightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=0\)

\(\Leftrightarrow2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=0\)

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

\(\Leftrightarrow\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}=|\frac{1}{x}+\frac{1}{y}+\frac{1}{z}|\)

\(x+y+z=0\)

\(\Leftrightarrow\frac{x+y+z}{xyz}=0\)(Vì \(x,y,z\ne0\))

\(\Leftrightarrow\frac{1}{yz}+\frac{1}{xz}+\frac{1}{xy}=0\)

\(\Leftrightarrow2\left(\frac{1}{yz}+\frac{1}{xz}+\frac{1}{xy}\right)=0\)

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

nên \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)

\(\Leftrightarrow\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}=\left|\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right|\)(Áp dụng HĐT \(\sqrt{x^2}=\left|x\right|\))

Áp dụng bđt côsi cho 2 số dương lần lượt ta có : 

\(1+\frac{y}{x}\ge2\sqrt{\frac{y}{x}}\)

\(1+\frac{z}{y}\ge2\sqrt{\frac{z}{y}}\)

\(1+\frac{x}{z}\ge2\sqrt{\frac{x}{z}}\)

Nhân vế theo vế ta đc : \(\left(1+\frac{y}{x}\right)\left(1+\frac{z}{y}\right)\left(1+\frac{x}{z}\right)\ge8\sqrt{\frac{xyz}{xyz}}=8\)

Dấu  = xảy ra khi : \(1=\frac{y}{x}\)=> x=y  và \(1=\frac{z}{y}\) => z=y và \(1=\frac{x}{z}\) => x=z

=> x=y=z

Thay vào M ta được : \(M=\frac{x^2}{2x^2}+\frac{y^2}{2y^2}+\frac{z^2}{2z^2}=\frac{3}{2}\).