Start by multiplying each side of the inequality by \(\sqrt{x + 1}\) and simplifying:
\(2 \sqrt x + \frac{1}{\sqrt{x+1}} \leq 2 \sqrt{x + 1}\)
\((2 \sqrt x + \frac{1}{\sqrt{x+1}})(\sqrt{x + 1}) \leq (2 \sqrt{x + 1})(\sqrt{x + 1})\)
\(2 \sqrt{x(x + 1)} + 1 \leq 2(x + 1)\)
\(2 \sqrt{x^2 + x} + 1 \leq 2x + 2\)
\(2 \sqrt{x^2 + x} \leq 2x + 1\)
\(\sqrt{x^2 + x} \leq x + \frac{1}{2}\)
From here, we can square both sides to get
\(x^2 + x \leq (x + \frac{1}{2})^2\)
\(x^2 + x \leq x^2 + x + \frac{1}{4}\)
\(0 \leq \frac{1}{4}\), which is always true.
