Ta có: \(a< a+b\left(a,b>0\right)\Rightarrow\frac{a}{a+b}< 1\)
Có: \(\frac{a}{a+b}=\sqrt{\frac{a}{a+b}}.\sqrt{\frac{a}{a+b}}\)
Lại có: \(\frac{a}{b+a}< 1\Leftrightarrow\sqrt{\frac{a}{b+a}}< 1\Rightarrow\sqrt{\frac{a}{a+b}}.\sqrt{\frac{a}{a+b}}< \sqrt{\frac{a}{a+b}}\Rightarrow\frac{a}{a+b}< \sqrt{\frac{a}{a+b}}\)
Chứng minh tương tự ta có:
\(\frac{b}{b+c}< \sqrt{\frac{b}{b+c}}\)
\(\frac{c}{c+a}< \sqrt{\frac{c}{c+a}}\)
\(\Rightarrow\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< \sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\)
đpcm
Sai thì thôi nhé~
Mới lp 8