Bài 184:Cho x>0 .CMR: $\frac{2\sqrt{2}}{\sqrt{x+1}}+\sqrt{x}\leq \sqrt{x+9}$
$BDT\Leftrightarrow (2\sqrt{2}+\sqrt{x(x+1)})^{2}\leq (x+1)(x+9)\Leftrightarrow 8+x^{2}+x+4\sqrt{2x(x+1)}\leq x^{2}+10x+9\Leftrightarrow 4\sqrt{2x(x+1)}\leq 9x+1=(x+1)+8x\Leftrightarrow (\sqrt{x+1}-2\sqrt{2x})^{2}\geq 0.$