Let x > 0, prove x^2 + 1 > lnx?

Jun 22, 2018

Two cases: $0 < x < 1$ and $x \ge 1$

For $x < 1$ the left side ${x}^{2} + 1$ is positive and the right side $\ln x$ is negative, so the statement is true.

For $x \ge 1$ consider

$f \left(x\right) = {x}^{2} + 1 - \ln x$

$f ' \left(x\right) = 2 x - \frac{1}{x}$

For all $x \ge 1$ we see $f ' \left(x\right) > 0$ so $f$ is increasing over the domain $x \ge 1.$ Since $f \left(1\right) = 2 > 0$ and $f$ is increasing, we have $f \left(x\right) > 0$ for all $x \ge 1$ or

${x}^{2} + 1 > \ln x$

We've shown both cases, so this is true for all $x > 0.$

Jun 22, 2018

As $\frac{d}{\mathrm{dx}} \ln x = \frac{1}{x}$ grows much slower than $\frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right) = 2 x$, it follows that ${x}^{2} + 1 > \ln x$

Explanation:

For any number $0 < x \le 1$ we have $\ln x = < 0$, so it's quite obvious that in this interval ${x}^{2} + 1 > \ln x$ in this interval.

$\frac{d}{\mathrm{dx}} \ln x = \frac{1}{x}$ (I won't prove it here - you can find the proof on the net.
As $\frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right) = 2 x$ we can see that ${x}^{2} + 1$ increases much faster than $\ln x$. It, therefore, follows that ${x}^{2} + 1 > \ln x$

Graphically it is quite clear: 