What kind of inequality can you generate from it?
It's obvious that for x>1, the graph of y=x−1 is always greater than y=lnx.
Therefore, we can say that lnx<x−1 for x>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 x2 such that we have the inequality now defined:
ln(x2)<x2−1 for x2>1
That is,
ln(x2)<x2−1 for x>2
Or we could also replace x by x+1 and get:
ln(x+1)<(x+1)−1 for x+1>1
which, after cleaning up a bit we get:
ln(x+1)<x for x>0
We can exploit this to prove the inequality below:
(1+1219)(1+1319)<e1218
From the above inequality formula where ln(x+1)<x for x>0, we have:
ln(1+1219)<1219 and
ln(1+1319)<1319
Adding both we obtain:
ln(1+1219)+ln(1+1319)<1219+1319
ln(1+1219)(1+1319)<1219+1319<1219+1219=2219=1218
Exponentiate both sides we get:
(1+1219)(1+1319)<e1218
and we're hence done! :D
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