What kind of inequality can you generate from it?

It's obvious that for $x\gt 1$, the graph of $y=x-1$ is always greater than $y=\ln x$.

Therefore, we can say that $\ln x\lt x-1$ for $x\gt 1$.

This is a very important discovery and this discovery can help us to prove many HARD and DIFFICULT IMO inequality problem.

But, do you also aware that we could replace $x$ by any suitable replacement we would like and use it to our advantage?

Take for example, we could replace $x$ by $\dfrac{x}{2}$ such that we have the inequality now defined:

$\ln \left(\dfrac{x}{2}\right)\lt \dfrac{x}{2}-1$ for $\dfrac{x}{2}\gt 1$

That is,

$\ln \left(\dfrac{x}{2}\right)\lt \dfrac{x}{2}-1$ for $x\gt 2$

Or we could also replace $x$ by $x+1$ and get:

$\ln (x+1)\lt (x+1)-1$ for $x+1\gt 1$

which, after cleaning up a bit we get:

$\ln (x+1)\lt x$ for $x\gt 0$

We can exploit this to prove the inequality below:

$\large \left(1+\dfrac{1}{2^{19}}\right)\left(1+\dfrac{1}{3^{19}}\right)\lt e^{\frac{1}{2^{18}}}$

From the above inequality formula where $\ln (x+1)\lt x$ for $x\gt 0$, we have:

$\ln \left(1+\dfrac{1}{2^{19}}\right)\lt \dfrac{1}{2^{19}}$ and

$\ln \left(1+\dfrac{1}{3^{19}}\right)\lt \dfrac{1}{3^{19}}$

Adding both we obtain:

$\ln \left(1+\dfrac{1}{2^{19}}\right)+\ln \left(1+\dfrac{1}{3^{19}}\right) \lt \dfrac{1}{2^{19}}+\dfrac{1}{3^{19}}$

$\ln \left(1+\dfrac{1}{2^{19}}\right)\left(1+\dfrac{1}{3^{19}}\right) \lt \dfrac{1}{2^{19}}+\dfrac{1}{3^{19}}\lt \dfrac{1}{2^{19}}+\dfrac{1}{2^{19}}=\dfrac{2}{2^{19}}=\dfrac{1}{2^{18}}$

Exponentiate both sides we get:

$\large \left(1+\dfrac{1}{2^{19}}\right)\left(1+\dfrac{1}{3^{19}}\right) \lt e^{\frac{1}{2^{18}}}$

and we're hence done! :D

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