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Compare the numbers:
$$\sqrt{x^2+1}$$ and $$x$$

Note by Oussama Jaber
3 years, 8 months ago

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I am assuming that the principle square root is taken. Note that $$x=\sqrt{x^2}$$. Obviously $$\sqrt{x^2+1}>\sqrt{x^2}$$; therefore $$\sqrt{x^2+1}>x$$.

CAUTION: observe that $$\sqrt{x^2}=|x|\ne x$$; however, even if $$x$$ is negative it does not invalidate the inequality. · 3 years, 8 months ago