Bài 1: Cho a,b,c $\geq$ 0 . Tìm GTNN của S=$\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b}+\frac{d}{a+b+c}+\frac{b+c+d}{a}+\frac{c+d+a}{b}+\frac{d+a+b}{c}+\frac{a+b+c}{d}$
Bài 2: Cho a,b,c $\geq$ 0 và $a^{2}+b^{2}+c^{2}= 1$ . Tìm GTNN của S=$a+b+c+\frac{1}{abc}$
1.$S=\sum (\frac{a}{b+c+d}+\frac{b+c+d}{9a})+\sum \frac{8}{9}.\frac{b+c+d}{a}\geq 8\sqrt[8]{\frac{abcd.\coprod (b+c+d)}{9^{4}.abcd.\prod (b+c+d)}}+\frac{8}{9}(\frac{a}{b}+ \frac{a}{c}+\frac{a}{d}+\frac{b}{c}+\frac{b}{a}+\frac{b}{d}+\frac{c}{a}+\frac{c}{b}+\frac{c}{d}+\frac{d}{c}+\frac{d}{a}+\frac{d}{b})$
$\Leftrightarrow S\geq \frac{8}{3}+\frac{8}{9}.12=\frac{40}{3}$
Dấu''='' xảy ra khi $a=b=c=d$
2.$S=a+b+c+\frac{1}{abc}=a+b+c+\frac{1}{9abc}+\frac{8}{9abc}\geq 4\sqrt[4]{a.b.c.\frac{1}{9abc}}+\frac{8}{9(\sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}})^{3}}=\frac{4}{\sqrt{3}}+\frac{8}{\sqrt{3}}=\frac{12}{\sqrt{3}}=4\sqrt{3}$
Dấu''='' xảy ra $\Leftrightarrow a=b=c=\frac{1}{\sqrt{3}}$