do x+y+z=1 nên 1/x+1/y+1/z sẽ bằng \(\frac{x+y+z}{x}+\frac{x+y+z}{y}+\frac{x+y+z}{z}=1+\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+1+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}+1\)

\(=3+\frac{y}{x}+\frac{x}{y}+\frac{y}{z}+\frac{z}{y}+\frac{x}{z}+\frac{z}{x}\)

Ta có

 \(\frac{x}{y}+\frac{y}{z}\ge2\)

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

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

Cộng vế theo vế của 3 bất đẳng thức trên ta được

\(\frac{y}{x}+\frac{x}{y}+\frac{y}{z}+\frac{z}{y}+\frac{x}{z}+\frac{z}{x}\ge6\)

Cộng 3 vào 2 vế bất đẳng thức 

\(\Rightarrow3+\frac{y}{x}+\frac{x}{y}+\frac{y}{z}+\frac{z}{y}+\frac{x}{z}+\frac{z}{x}\ge9\)

Mà \(3+\frac{y}{x}+\frac{x}{y}+\frac{y}{z}+\frac{z}{y}+\frac{x}{z}+\frac{z}{x}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge9\)

Xong !!!!

T I C K nha cảm ơn nhìu

CHÚC BẠN HỌC TỐT

Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có ngay :

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{\left(1+1+1\right)^2}{x+y+z}=\frac{9}{x+y+z}=9\left(đpcm\right)\)

Dấu "=" xảy ra <=> x=y=z=1/3

Ta có:

\(x^3+y^3+z^3=3xyz\)

nên  \(x^3+y^3+z^3-3xyz=0\)

\(\Leftrightarrow\left(x^3+y^3\right)+z^3-3xyz=0\)

\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)

\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2+\left(x+y\right).z+z^2\right]-3xy\left(x+y+z\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2-3xy\right]=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=0\)

\(\Leftrightarrow\frac{1}{2}\left(x+y+z\right)\left(2x^2+2y^2+2z^2-2xy-2xz-2yz\right)=0\)

\(\Leftrightarrow\frac{1}{2}\left(x+y+z\right)\left[\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2xz+x^2\right)\right]=0\)

\(\Leftrightarrow\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)

\(\Leftrightarrow^{x+y+z=0}_{x=y=z}\)

Do đó:

\(M=\left(2-\frac{x}{y}\right)^{2013}+\left(3-\frac{2x}{z}\right)^{2014}+\left(4-\frac{3z}{x}\right)^{2015}\)

\(=\left(2-\frac{y}{y}\right)^{2013}+\left(3-\frac{2z}{z}\right)^{2014}+\left(4-\frac{3x}{x}\right)^{2015}\)

\(=\left(2-1\right)^{2013}+\left(3-2\right)^{2014}+\left(4-3\right)^{2015}\)

\(M=1^{2013}+1^{2014}+1^{2015}=1+1+1=3\)

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