Question

cho a,b,c >0 chứng minh rằng \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2\sqrt[3]{abc}}>=\frac{\left(a+b+c+\sqrt[3]{abc}\right)^{ }}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\) 

Answer 1

\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2\sqrt[3]{abc}}\ge\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

cauchy-schwarz: 

\(VT=\frac{c^2}{ac^2+bc^2}+\frac{a^2}{a^2b+a^2c}+\frac{b^2}{b^2c+b^2a}+\frac{\sqrt[3]{a^2b^2c^2}}{2abc}\ge\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)