Inequality proof.

**Also sprach Zarathustra** · June 28th 2011, 04:03 AM

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

**girdav** · June 28th 2011, 08:15 AM

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.

**Also sprach Zarathustra** · June 28th 2011, 08:59 PM

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.

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