Question

Cho a,b,c > 0. Chứng minh

∛\(\left(\dfrac{2a}{b+c}\right)^2\) +  ∛\(\left(\dfrac{2b}{c+a}\right)^2\)  +   ∛\(\left(\dfrac{2c}{a+b}\right)^2\)   ≥ 3

Answer 1

\(\sqrt[3]{\left(\dfrac{2a}{b+c}\right)^2}=\dfrac{1}{\sqrt[3]{\dfrac{b+c}{2a}\cdot\dfrac{b+c}{2a}\cdot1}}>=\dfrac{3}{\dfrac{b+c}{2a}+\dfrac{b+c}{2a}+1}=\dfrac{3a}{a+b+c}\)

Chứng minh tương tự, ta được:

\(\sqrt[3]{\left(\dfrac{2b}{a+c}\right)^2}>=\dfrac{3b}{a+b+c}\)

và \(\sqrt[3]{\left(\dfrac{2c}{a+b}\right)^2}>=\dfrac{3c}{a+b+c}\)

Cộng vế theo vế, ta được:

\(\sqrt[3]{\left(\dfrac{2a}{b+c}\right)^2}+\sqrt[3]{\left(\dfrac{2b}{a+c}\right)^2}+\sqrt[3]{\left(\dfrac{2c}{a+b}\right)^2}>=\dfrac{3a+3b+3c}{a+b+c}=3\)