Use the mean value theorem to show that
$$
1-\frac{1}{\sqrt{1+x^{2}}} \leq x^{2}, \quad x>0
$$
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If $f(x)=\left(1+x^{2}\right)^{-1 / 2}$, then, $f(x)-f(0)=f^{\prime}(c)(x-0)$ with $0<c<x$. Since $f^{\prime}(c)=-c\left(1+c^{2}\right)^{-3 / 2}>-c>-x$ for $x>0$,
$$
\frac{1}{\sqrt{1+x^{2}}}-1=f^{\prime}(c) x \geq-x^{2}
$$
which implies the result.
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