Question

Cho a , b ,c >0 . cmr: \(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)

Answer 1

Theo bất đẳng thức cô si, có:

\(\sqrt{1.\dfrac{b+c}{a}}\le\left(1+\dfrac{b+c}{a}\right):2=\dfrac{a+b+c}{2a}\)

\(\Rightarrow\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}~~~~~\left(1\right)\)

Tương tự: \(\sqrt{\dfrac{b}{c+a}}\ge\dfrac{2b}{a+b+c}~~~~~\left(2\right)\)

\(\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}~~~~~\left(3\right)\)

Cộng vế theo vế \(\left(1\right);\left(2\right);\left(3\right)\), ta có:

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