ĐKXĐ: \(x\ge1\)
Ta có:
\(\sqrt{x-1}=2.\frac{1}{2}\sqrt{x-1}\le\frac{1}{4}+x-1=x-\frac{3}{4}\)
\(9\sqrt{x+1}=6.\frac{3}{2}.\sqrt{x+1}\le3\left(\frac{9}{4}+x+1\right)=3x+\frac{39}{4}\)
\(\Rightarrow\sqrt{x-1}+9\sqrt{x+1}\le4x+9\)
Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}\sqrt{x-1}=\frac{1}{2}\\\sqrt{x+1}=\frac{3}{2}\end{matrix}\right.\) \(\Leftrightarrow x=\frac{5}{4}\)