Áp dụng bđt cosi vs hai số dương có
\(\sqrt{\frac{b+c}{a}}\le\frac{\frac{b+c}{a}+1}{2}=\frac{\frac{b+c+a}{a}}{2}=\frac{a+b+c}{2a}\)
=> \(\sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}\)
Tương tự cx có: \(\sqrt{\frac{b}{a+c}}\ge\frac{2b}{a+b+c}\)
\(\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\)
=> \(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge\frac{2a}{a+b+c}+\frac{2b}{a+b+c}+\frac{2c}{a+b+c}\)
<=> \(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge\frac{2a+2b+2c}{a+b+c}=2\)
Dấu "=" xảy ra <=> a+b+c=0( k xảy ra)
=> dấu "=" k xảy ra
Vậy \(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}>2\)
AD BDT AM-GM ta có:
\(a+b+c\ge2\sqrt{a\left(b+c\right)}\)
\(\Leftrightarrow\frac{1}{2\sqrt{a\left(b+c\right)}}\ge\frac{1}{a+b+c}\)
\(\Leftrightarrow\sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}\)
CM tương tự
\(\Rightarrow\sqrt{\frac{b}{a+c}}\ge\frac{2b}{a+b+c};\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\)
\(\Rightarrow\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge\frac{2\left(a+b+c\right)}{a+b+c}=2\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a=b+c\\b=c+a\\c=a+b\end{matrix}\right.\) \(\Leftrightarrow a=b=c=0\) (trái vs gt)
Vậy dấu "=" ko xảy ra
\(\Rightarrow\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}>2\left(dpcm\right)\)
