namtuoc

$BĐT<=>$\sum \sqrt[4]{a^{3}}>\sqrt[4]{(a+b+c)^{3}} <=>\sum \sqrt[4]{\frac{a}{a+b+c}}>1.Đúng do \frac{a}{a+b+c};\frac{b}{a+b+c};\frac{c}{a+b+c}<1 nên \sum \sqrt[4]{\frac{a}{a+b+c}}<\frac{a+b+c}{a+b+c}=1.=>Q.E.D$