Đề:  Cho a,b,c dương. Tìm GTNN $T= \frac{a}{b+c+d}+\frac{b+c+d}{a}+\frac{b}{c+d+a}+\frac{c+d+a}{b}+\frac{c}{d+a+b}+\frac{d+a+b}{c}+\frac{d}{a+b+c}+\frac{a+b+c}{d}   $

Đề bài:  Cho a,b,c dương. Tìm GTNN $T= \frac{a}{b+c+d}+\frac{b+c+d}{a}+\frac{b}{c+d+a}+\frac{c+d+a}{b}+\frac{c}{d+a+b}+\frac{d+a+b}{c}+\frac{d}{a+b+c}+\frac{a+b+c}{d}   $

Lời giải

$T=\frac{1}{3}[(\frac{3a}{b+c+d}+\frac{b+c+d}{3a}+(\frac{3b}{c+d+a}+\frac{c+d+a}{3b} )+(\frac{3c}{d+a+b}+\frac{d+a+b}{3c})+(\frac{3d}{a+b+c}+\frac{a+b+c}{3d})] $
              $+\frac{8}{9}[\frac{b+c+d}{a}+\frac{c+d+a}{b}+\frac{d+a+b}{c}+\frac{a+b+c}{d}]   $
     $\geq \frac{1}{3}[2+2+2+2]+\frac{8}{9}[(\frac{b}{a}+\frac{a}{b})+(\frac{c}{a}+\frac{a}{c})+(\frac{d}{a}+\frac{a}{d})+(\frac{c}{b}+\frac{b}{c})+(\frac{d}{b}+\frac{b}{d})$
          $+(\frac{d}{c}+\frac{c}{d} )     $
     $ \geq  \frac{8}{3}+\frac{8}{9}[2+2+2+2+2+2]=\frac{40}{3}   $
Vậymin$T=\frac{40}{3} $ khi  $a=b=c=d$