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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