Với a, b > 0
Ta có: \(2\sqrt{a+3}\le\frac{\left(a+3\right)+4}{2}\)
\(\Leftrightarrow2\sqrt{a+3}\le\frac{a+2}{2}\)
\(\Leftrightarrow\frac{2}{\sqrt{a+3}}\ge\frac{8}{a+7}\)
Ta có: \(2\sqrt{b+3}\le\frac{\left(b+3\right)+4}{2}\)
\(\Leftrightarrow\frac{1}{\sqrt{b+3}}\ge\frac{4}{b+7}\)
\(\Rightarrow\frac{2}{\sqrt{a+3}}+\frac{1}{\sqrt{b+3}}\ge\frac{8}{a+7}+\frac{4}{b+7}=\frac{4}{a+7}+\frac{4}{a+7}+\frac{4}{b+7}\)
\(\ge4\left(\frac{1}{a+7}+\frac{1}{a+7}+\frac{1}{b+7}\right)\)
\(\ge4.\frac{9}{2a+b+21}=4.\frac{9}{3+21}=\frac{36}{24}\)
\(\ge\frac{3}{2}\left(đpcm\right)\)
Vậy\(\frac{2}{\sqrt{a+3}}+\frac{1}{\sqrt{b+3}}\ge\frac{3}{2}\)
Cách khác:
Ta có: \(VT=\frac{2}{\sqrt{a+3}}+\frac{1}{\sqrt{b+3}}=\frac{2}{\sqrt{\left(a+1\right)+2}}+\frac{1}{\sqrt{\left(b+1\right)+2}}\ge\frac{2}{\frac{a+1+2}{2}}+\frac{1}{\frac{b+1+2}{2}}=\frac{4}{a+3}+\frac{2}{b+3}\)(1) (BĐT Cô-si)
Lại có: \(2a+b\le3\Leftrightarrow\left\{{}\begin{matrix}a+3\ge3a+b\\b+3\ge2\left(a+b\right)\end{matrix}\right.\). Thay vào (1) ta được:
\(VT\ge\frac{4}{3a+b}+\frac{1}{a+b}\)
Áp dụng BĐT Schwarz, ta được:
\(VT\ge\frac{4}{3a+b}+\frac{1}{a+b}\ge\frac{\left(2+1\right)^2}{4a+2b}=\frac{3^2}{2\left(2a+b\right)}\ge\frac{3^2}{2.3}=\frac{3}{2}\)(đpcm)
Dấu "=" xảy ra khi và chỉ khi a=b=1
