Question

Cho a,b,c dương. Chứng minh: \(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)

Answer 1

Áp dụng bđt AM-GM:

\(\left\{{}\begin{matrix}\sqrt{a\left(b+c\right)}\le\dfrac{a+b+c}{2}\\\sqrt{b\left(c+a\right)}\le\dfrac{a+b+c}{2}\\\sqrt{c\left(a+b\right)}\le\dfrac{a+b+c}{2}\end{matrix}\right.\)

Ta có: \(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{c}{a+b}}\)

\(=\dfrac{a}{\sqrt{a\left(b+c\right)}}+\dfrac{b}{\sqrt{b\left(a+c\right)}}+\dfrac{c}{\sqrt{c\left(a+b\right)}}\)

\(\ge\dfrac{a}{\dfrac{a+b+c}{2}}+\dfrac{b}{\dfrac{a+b+c}{2}}+\dfrac{c}{\dfrac{a+b+c}{2}}\)

\(=\dfrac{2\left(a+b+c\right)}{\left(a+b+c\right)}=2\)

Dấu "=" ko xảy ra nên ta có đpcm