Question

cho a,b,c là các số dương . Tìm GTNN của bt; Q=\(\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\)

Answer 1

+ \(\sqrt{4a\left(3a+b\right)}\le\frac{4a+3a+b}{2}=\frac{7a+b}{2}\)

Dấu "=" \(\Leftrightarrow4a=3a+b\Leftrightarrow a=b\)

+ \(\sqrt{4b\left(3b+a\right)}\le\frac{a+7b}{2}\) Dấu "=" \(\Leftrightarrow a=b\)

\(\Rightarrow\sqrt{4a\left(3a+b\right)}+\sqrt{4b\left(3b+a\right)}\le4\left(a+b\right)\)

\(\Rightarrow\frac{1}{2}Q=\frac{a+b}{\sqrt{4a\left(3a+b\right)}+\sqrt{4b\left(3b+a\right)}}\ge\frac{a+b}{4\left(a+b\right)}=\frac{1}{4}\)

\(\Rightarrow Q\ge\frac{1}{2}\)

Dấu "=" \(\Leftrightarrow a=b\)