\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)

\(\Leftrightarrow\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\ge4\)

Áp dụng BĐT Cô - si

\(\Rightarrow\left\{\begin{matrix}x+y\ge2\sqrt{xy}\\\frac{1}{x}+\frac{1}{y}\ge2\sqrt{\frac{1}{xy}}\end{matrix}\right.\)

\(\Rightarrow\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\ge4\sqrt{xy.\frac{1}{xy}}\)

\(\Rightarrow\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\ge4\) ( đpcm )

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)

\(\Leftrightarrow\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\)

Áp dụng BĐT Cô - si

\(\Rightarrow\left\{\begin{matrix}x+y+z\ge3\sqrt{xyz}\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt{\frac{1}{xyz}}\end{matrix}\right.\)

\(\Rightarrow\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\sqrt{xyz.\frac{1}{xyz}}\)

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

từ đề bài ta có bất đẳng thức cần chứng minh tương đương: 

\(3+\dfrac{z}{x+y}+\dfrac{x}{y+z}+\dfrac{y}{x+z}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{9}{4}\)

<=>\(\dfrac{3}{4}+\dfrac{z}{x+y}+\dfrac{x}{y+z}+\dfrac{y}{x+z}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

ta có \(\dfrac{3}{4}+\dfrac{z}{x+y}+\dfrac{x}{y+z}+\dfrac{y}{x+z}\le\dfrac{3}{4}+\dfrac{z+y}{4x}+\dfrac{x+z}{4y}+\dfrac{x+y}{4z}=\dfrac{3}{4}+\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{3}{4}=\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\left(đpcm\right)\)Dấu "=" xảy ra khi x=y=z=\(\dfrac{1}{3}\)

Áp dụng bất đẳng thức Bunhiacopxki, ta có : \(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge\left(1+1+1\right)^2=9\)

Áp dụng cô si 3 số dương:

\(x+y+z\ge3\sqrt[3]{xyz}\)

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt[3]{\frac{1}{xyz}}\)

Nhân lại theo từng vế:\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\sqrt[3]{xyz\times\frac{1}{xyz}}=9\times1=9\)(Đpcm)

bài này bạn thêm x,y,z dương nx nhé

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

Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)

\(\sqrt{2x\left(y+z\right)}< =\dfrac{2x+y+z}{2}\)

=>\(\dfrac{1}{\sqrt{x\left(y+z\right)}}>=\dfrac{2\sqrt{2}}{2x+y+z}\)

=>\(P>=2\sqrt{2}\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\right)\)

\(\Leftrightarrow P>=2\sqrt{2}\cdot\dfrac{\left(1+1+1\right)^2}{\left(2x+y+z\right)+x+2y+z+x+y+2z}=\dfrac{18\sqrt{2}}{4\cdot18\sqrt{2}}=\dfrac{1}{4}\)

Dấu = xảy ra khi x=y=z=6căn 2