Question

Cho a,b,c > 0 . Chứng minh rằng 

\(\sqrt{\frac{a}{b+c}}\:+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\:>2\)>2

Ai trl đc mk tick chi nha :)))

Answer 1

Ta có: \(\sqrt{\frac{b+c}{a}}\le\frac{1+\frac{b+c}{a}}{2}=\frac{a+b+c}{2a}\) 

     \(\Rightarrow\sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}\)  

Tương tự \(\sqrt{\frac{b}{c+a}}\ge\frac{2b}{a+b+c};\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\) 

\(\Rightarrow\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\ge2\) 

Dấu "=" xảy ra khi a=b=c=0 (trái gt) 

\(\Rightarrow\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}>2\)(đpcm)