Question

Cho x,y,z>0, \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\) CMR

\(\dfrac{x^2}{x+yz}+\dfrac{y^2}{y+zx}+\dfrac{z^2}{z+xy}\ge\dfrac{x+y+z}{4}\)

Answer 1

Ta có \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\Rightarrow ab+bc+ca=abc\)

Có: \(\dfrac{a^2}{a+bc}=\dfrac{a^3}{a^2+abc}=\dfrac{a^3}{a^2+ab+ac+bc}=\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}\)

Áp dụng bđt cô si cho 3 số dương ta có: 

\(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{a^3\left(a+b\right)\left(a+c\right)}{\left(a+b\right)\left(a+c\right)8.8}}=\dfrac{3a}{4}\)

Cmtt: ....

Cộng vế ta được:

(Đpcm)

Dấu = xảy ra khi a=b=c=3