Tuesday, September 1, 2015

Develep the power of observation

If you're given the two graphs that are plotted in the same Cartesian plane below, what do you notice?

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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