Inequality proof.

• Jun 28th 2011, 04:03 AM
Also sprach Zarathustra
Inequality proof.
Prove that $\displaystyle (x+1)\ln(x+1)+e^y\geq (1+x)(1+y)$, where $\displaystyle x,y \in \mathbb{R}^+$
• Jun 28th 2011, 08:15 AM
girdav
Re: Inequality proof.
We fix $\displaystyle x\geq 0$. Let $\displaystyle f(y) := (x+1)\ln(x+1)+e^y-(1+x)(1+y)$. We have $\displaystyle f'(y) =e^y-(1+x)$ hence $\displaystyle f$ is decreasing on $\displaystyle \left[0,\ln(1+x)\right]$ and increasing on $\displaystyle \left[\ln(1+x),+\infty[$. We get that $\displaystyle f$ gets its minimum at $\displaystyle y =\ln (1+x)$.
Since \displaystyle \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.
• Jun 28th 2011, 08:59 PM
Also sprach Zarathustra
Re: Inequality proof.
Quote:

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

(Clapping)

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