\(S=\sqrt{x-3}+\sqrt{y-4}\)
ĐK:\(x\ge 3;y\ge 4\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(S^2=\left(\sqrt{x-3}+\sqrt{y-4}\right)^2\)
\(\le\left(1+1\right)\left(x-3+y-4\right)\)
\(=2\left(x+y-7\right)=2\)
\(\Rightarrow S^2\le2\Rightarrow S\le\sqrt{2}\)