An inequality problem 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.
 A: 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.
A: 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.
A: 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.
