Question

C/m\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{c}{a+b}}>2\left(a,b,c>0\right)\)

Answer 1

Vì a,b,c>0 áp dụng BĐT Cauchy ta có

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

Tương tự \(\sqrt{\dfrac{b}{c+a}}\ge\dfrac{2b}{a+b+c}\) (2) và \(\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}\) (3)

Cộng (1), (2), (3) vế theo vế

\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}\)

\(\ge2\left(\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}\right)=2\)

Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a=b+c\\b=c+a\\a=b+c\end{matrix}\right.\) \(\Leftrightarrow a=b=c=0\) (vô lý)

Vậy đẳng thức ko xảy ra, hay \(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)