\(\sqrt{x+1}+2\sqrt{2x+3}\ge2x+2\)
\(\Leftrightarrow\sqrt{x+1}-2+2\sqrt{2x+3}-6-2x+6\ge0\)
\(\Leftrightarrow\frac{x+1-4}{\sqrt{x+1}+2}+\frac{2\cdot\left(2x+3-9\right)}{\sqrt{2x+3}+3}-2\left(x-3\right)\ge0\)
\(\Leftrightarrow\frac{x-3}{\sqrt{x+1}+2}+\frac{4\cdot\left(x-3\right)}{\sqrt{2x+3}+3}-2\left(x-3\right)\ge0\)
\(\Leftrightarrow\left(x-3\right)\cdot\left(\frac{1}{\sqrt{x+1}+2}+\frac{4}{\sqrt{2x+3}+3}-2\right)\ge0\)
Xét \(\frac{1}{\sqrt{x+1}+2}+\frac{4}{\sqrt{2x+3}+3}-2=\frac{\sqrt{x+\frac{3}{2}}\cdot\sqrt{x+1}+\sqrt{\frac{x+1}{2}}+\frac{3}{2}\sqrt{x+\frac{3}{2}}+\frac{1}{2\sqrt{2}}}{\left(\sqrt{x+1}+2\right)\left(\sqrt{2x+3}+3\right)}\ge0\)
Do đó \(x-3\ge0\Leftrightarrow x\ge3\)
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