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Inequality proof.
Prove that \ln(x+1)+e^y\geq (1+x)(1+y))
, where 
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Re: Inequality proof.
We fix 
. Let  := (x+1)\ln(x+1)+e^y-(1+x)(1+y))
. We have  =e^y-(1+x))
hence 
is decreasing on ![\left[0,\ln(1+x)\right]](http://latex.codecogs.com/png.latex?\left[0,\ln(1+x)\right])
and increasing on ,+\infty[)
. We get that gets its minimum at )
.
Since ) &= (x+1)\ln(1+x)+1+x-(1+x)(1+\ln(1+x)) \\&= (1+x)(\ln(1+x)+1-(1+\ln(1+x)))\\& =0\end{align*})
the result is shown.
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Re: Inequality proof.
Quote:
Originally Posted by
girdav 
We fix
. Let
. We have
hence
is decreasing on
and increasing on
. We get that
gets its minimum at
.
Since
the result is shown.
(Clapping)
"My way":
