1. ## Inequality proof.

Prove that $(x+1)\ln(x+1)+e^y\geq (1+x)(1+y)$, where $x,y \in \mathbb{R}^+$

2. ## Re: Inequality proof.

We fix $x\geq 0$. Let $f(y) := (x+1)\ln(x+1)+e^y-(1+x)(1+y)$. We have $f'(y) =e^y-(1+x)$ hence $f$ is decreasing on $\left[0,\ln(1+x)\right]$ and increasing on $\left[\ln(1+x),+\infty[$. We get that $f$ gets its minimum at $y =\ln (1+x)$.
Since \begin{align*}f(\ln(1+x)) &= (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.

3. ## Re: Inequality proof.

Originally Posted by girdav
We fix $x\geq 0$. Let $f(y) := (x+1)\ln(x+1)+e^y-(1+x)(1+y)$. We have $f'(y) =e^y-(1+x)$ hence $f$ is decreasing on $\left[0,\ln(1+x)\right]$ and increasing on $\left[\ln(1+x),+\infty[$. We get that $f$ gets its minimum at $y =\ln (1+x)$.
Since \begin{align*}f(\ln(1+x)) &= (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.

"My way":
Spoiler:
Take the function $f(t)=\ln(t+1)$ and use Young's inequality - Wikipedia, the free encyclopedia