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I can't work out that inequality problem. If $x\ge0$, $y\ge0$, how to prove that $$ 1+x+y+xy\leq(x+1)\ln(x+1)+e^y? $$ I tried taylor expansions for $$ln(x+1)$$ and $$e^y,$$ I also tried $$ 1+x+y+xy=(1+x)(1+y),$$ but I still can't work it out.

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Hi, welcome to math.SE. Since you're new here, allow me to mention that people will be much more able and willing to help you if you tells us what you have tried already. That being said, have you tried using the Taylor expansions for $\ln(x+1)$ and $e^y$? –  Johannes Kloos May 23 '12 at 13:15
    
Yes, but it won't help a lot because there is x*y in the left side. Using Taylor expansions, we only get functions of x and y separately. The x*y in the left side is really annoying and I don't know how to deal with it. –  Pan Yan May 23 '12 at 13:19
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By the way, please tell us if this is homework or where you encountered it. For dealing with the $xy$, I would propose making a case distinction $x \ge y$, $x < y$ - this allows you to estimate $xy$ by either $x^2$ or $y^2$, which should make things easier. –  Johannes Kloos May 23 '12 at 13:20
    
I encountered it in a book. I tried 1+x+y+xy=(1+x)(1+y), but didn't know whether it helps. –  Pan Yan May 23 '12 at 13:24
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3 Answers 3

up vote 1 down vote accepted

For fixed $x$, look at the difference $f_x(y) = e^y + (x+1)\ln(x+1) - 1 - x - y - xy$. $f_x'(y) = e^y -1 - x$, and $f_x''(y) = e^y > 0$. Thus $f_x(y)$ has a global minimum where the first derivative is zero, namely at $y = \ln(1 + x)$. For this value of $y$ we have $$f_x(y) = 1 + x + (x+1)\ln(1+x) - 1 - x - \ln(1 + x) - x\ln(1 + x)$$ $$ = 0$$ Thus the difference is always nonnegative, and therefore the left-hand side of the original equation is bounded by the right-hand side as needed.

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Thank you very much!! –  Pan Yan May 23 '12 at 14:05
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Let $f(s)=e^s$. Since $f$ is convex then it have Legendre transformation. And this transformation is $f^*(t)=t\ln(t/e)$. Then from the very definition of Legendre transformation $$ st\leq f(s)+f^*(t)=e^s+t\ln(t/e) $$ Now we set, $s = x + 1$ and $t=e(y+1)$ and get the desired inequality.

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Thank you very much. –  Pan Yan May 23 '12 at 13:55
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Let $x=e^z-1$. The condition can be simplified to $z(1+y-e^z) \le e^y$. When z exceeds y, this becomes trivially true as LHS becomes negative. Else, $z(1+y-e^z) \le z(1+y-(1+z)) = z(y-z)$. Consider $f(y)=e^y-yz+z^2$. $f'(y)=e^y-z$ which is always positive for z < y. Even at y=z, $f(y)=e^y$ is positive. So f(y) remains positive, which proves our inequality.

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