Question

Cho a,b,c là các số dương,chứng minh rằng:

\(\frac{ab}{a+b+2c}+\frac{bc}{b+c+2a}+\frac{ca}{c+a+2b}\ge\frac{a+b+c}{4}\)

 

Answer 1

#)Giải :

Ta có : 

\(\hept{\begin{cases}\frac{ab}{b+c+a+b}\le\frac{ab}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\\\frac{bc}{a+b+a+c}\le\frac{bc}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\\\frac{ac}{b+c+a+b}\le\frac{ac}{4}\left(\frac{1}{b+c}+\frac{1}{a+b}\right)\end{cases}}\)

\(\Rightarrow VT\le\frac{1}{a+b}.\left(\frac{bc}{4}+\frac{ac}{4}\right)+\frac{1}{a+c}.\left(\frac{bc}{4}+\frac{ab}{4}\right)+\frac{1}{b+c}.\left(\frac{ac}{4}+\frac{ab}{4}\right)\)

\(=\frac{1}{a+b}.\frac{c\left(a+b\right)}{4}+\frac{1}{a+c}.\frac{b\left(a+c\right)}{4}+\frac{1}{b+c}.\frac{a\left(b+c\right)}{4}\)

\(=\frac{c}{4}+\frac{b}{4}+\frac{a}{4}\)

\(\Rightarrow\frac{a+b+c}{4}\)

\(\Rightarrowđpcm\)