Question

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

Answer 1

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(x) = x^2 + 1 - ln x`

`f'(x) = 2x - 1/x`

For all `x ge 1` we see `f'(x) > 0` so `f` is increasing over the domain `x ge 1.` Since `f(1)=2>0` and `f` is increasing, we have `f(x)>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.`

Answer 2

As `d/dx lnx=1/x` grows much slower than `d/dx (x^2+1)=2x`, it follows that `x^2+1>lnx`

Explanation:

For any number `0 < x <= 1` we have `lnx=<0`, so it's quite obvious that in this interval `x^2+1>lnx` in this interval.

`d/dx lnx=1/x` (I won't prove it here - you can find the proof on the net.
As `d/dx (x^2+1)=2x` we can see that `x^2+1` increases much faster than `lnx`. It, therefore, follows that `x^2+1>lnx`

Graphically it is quite clear: