Respuesta :

Beanie4185
Start by multiplying each side of the inequality by [tex]\sqrt{x + 1}[/tex] and simplifying:

[tex]2 \sqrt x + \frac{1}{\sqrt{x+1}} \leq 2 \sqrt{x + 1}[/tex]
[tex](2 \sqrt x + \frac{1}{\sqrt{x+1}})(\sqrt{x + 1}) \leq (2 \sqrt{x + 1})(\sqrt{x + 1})[/tex]
[tex]2 \sqrt{x(x + 1)} + 1 \leq 2(x + 1)[/tex]
[tex]2 \sqrt{x^2 + x} + 1 \leq 2x + 2[/tex]
[tex]2 \sqrt{x^2 + x} \leq 2x + 1[/tex]
[tex]\sqrt{x^2 + x} \leq x + \frac{1}{2}[/tex]

From here, we can square both sides to get

[tex]x^2 + x \leq (x + \frac{1}{2})^2[/tex]
[tex]x^2 + x \leq x^2 + x + \frac{1}{4}[/tex]
[tex]0 \leq \frac{1}{4}[/tex], which is always true.

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