Áp dụng bất đẳng thức Cô - si (a, b, c) dương ta có:
+/ \(\sqrt{\dfrac{b+c}{a}}=\sqrt{\dfrac{b+c}{a}.1}\le(\dfrac{b+c}{a}+1):2=\dfrac{a+b+c}{2a}\)
\(\Rightarrow\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)
+/ \(\sqrt{\dfrac{a+c}{b}}=\sqrt{\dfrac{a+c}{b}.1}\le(\dfrac{a+c}{b}+1):2=\dfrac{a+b+c}{2b}\)
\(\Rightarrow\sqrt{\dfrac{b}{a+c}}\ge\dfrac{2b}{a+b+c}\)+/ \(\sqrt{\dfrac{a+b}{c}}=\sqrt{\dfrac{a+b}{c}.1}\le(\dfrac{a+b}{c}+1):2=\dfrac{a+b+c}{2c}\)
\(\Rightarrow\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}\)
Khi đó:\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2a+2b+2c}{a+b+c}=2\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a=b+c\\b=a+c\\c=a+b\end{matrix}\right.\)
\(\Leftrightarrow a+b+c=0\) (Trái với giả thiết a, b, c là 3 số dương)
=> Đẳng thức không xảy ra
=> \(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{c}{a+b}}>2\) (đpcm)
