Question

\(\frac{a+b+c}{\sqrt{a\left(a+3b\right)+\sqrt{b\left(b+3c\right)}+\sqrt{c\left(c+3c\right)}}}\ge\frac{1}{2}\)

Answer 1

Ta có : \(\frac{a+b+c}{\sqrt{a\left(a+3b\right)}+\sqrt{b\left(b+3c\right)}+\sqrt{c\left(c+3a\right)}}=\frac{2\left(a+b+c\right)}{\sqrt{4a\left(a+3b\right)+\sqrt{4b\left(b+3c\right)}+\sqrt{4c\left(c+3a\right)}}}\)

Áp dụng BĐT Cauchy , ta có :

\(\frac{2\left(a+b+c\right)}{\sqrt{4a\left(a+3b\right)}+\sqrt{4b\left(b+3c\right)}+\sqrt{4c\left(c+3a\right)}}\le\frac{2\left(a+b+c\right)}{\frac{4a+a+3b}{2}+\frac{4b+b+3c}{2}+\frac{4c+c+3a}{2}}=\frac{2\left(a+b+c\right)}{4\left(a+b+c\right)}=\frac{1}{2}\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)