# An Inequality in Four Weighted Variables

### Solution 1

The inequality is equivalent to

$\displaystyle \left(\frac{a+b+c+d}{c+d}\right)^{c+d}\ge\left(\frac{a+c}{c}\right)^c\left(\frac{b+d}{d}\right)^d,$

or,

$\displaystyle \left(\frac{a+b+c+d}{c+d}\right)^{c+d}\ge\left(\frac{a}{c}+1\right)^c\left(\frac{b}{d}+1\right)^d.$

Now, by the weighted AM-GM inequality,

$\displaystyle \frac{\displaystyle \left(\frac{a}{c}+1\right)c+\left(\frac{b}{d}+1\right)d}{\displaystyle c+d}\ge\sqrt[c+d]{\left(\frac{a}{c}+1\right)^c\left(\frac{b}{d}+1\right)^d},$

i.e.,

$\displaystyle \left(\frac{a+b+c+d}{c+d}\right)^{c+d}\ge\left(\frac{a}{c}+1\right)^c\left(\frac{b}{d}+1\right)^d.$

QED.

### Solution 2

Let $p=a+c$ and $q=b+d$. Noting that the $log$ function is concave, Jensen's inequality implies

$\displaystyle \frac{c}{c+d}\log\left(\frac{p}{c}\right)+\frac{d}{c+d}\log\left(\frac{q}{d}\right) \leq \log\left(\frac{p+q}{c+d}\right).$

The arguments lie in the domain of the $\log$ function as $p\gt c$ and $q\gt d$. Applying the monotonically increasing exponential function to both sides,

\displaystyle \begin{align} \left(\frac{p}{c}\right)^{c/(c+d)} \left(\frac{q}{d}\right)^{d/(c+d)} &\leq \left(\frac{p+q}{c+d}\right) \\ \Longrightarrow \left(\frac{p}{c}\right)^c \left(\frac{q}{d}\right)^d &\leq \left(\frac{p+q}{c+d}\right)^{c+d} \\ \Longrightarrow p^c q^d (c+d)^{c+d} &\leq c^c d^d (p+q)^{c+d} \\ \Longrightarrow (a+c)^c (b+d)^d (c+d)^{c+d} &\leq c^c d^d (a+b+c+d)^{c+d} \end{align}

### Solution 3

Define $f:\,(0,\infty)\to \mathbb{R},\,$ with $f(x)=\log (x+1).\,$ We have, $\displaystyle f'(x)=\frac{1}{x+1};,\,$ $\displaystyle f''(x)=\frac{-1}{(x+1)^2}\lt 0;\,$ so that $f\,$ is concave. By Jensen's inequality,

\displaystyle \begin{align} c \log \Bigr(1+\frac{a}{c}\Bigr)+d\log \Bigr(1+\frac{b}{d}\Bigr)&= (c+d)\Bigr(\frac{c}{c+d}\log \Bigr(1+\frac{a}{c}\Bigr)+\frac{d}{c+d}\log \Bigr(1+\frac{b}{d}\Bigr)\Bigr)\\ &\leq (c+d)\log \Bigr(1+\frac{c}{c+d}\cdot \frac{a}{c}+\frac{d}{c+d}\cdot \frac{b}{d}\Bigr)\\ &= (c+d)\log \Bigr(1+\frac{a}{c+d}+\frac{b}{c+d}\Bigr)\\ &=\log \Bigr(1+\frac{a+b}{c+d}\Bigr)^{c+d}. \end{align}

Further

\displaystyle \begin{align} &\log \Bigr[\Bigr(1+\frac{a}{c}\Bigr)^c\cdot \Bigr(1+\frac{b}{d}\Bigr)^d\Bigr]\leq \log \Bigr(\frac{a+b+c+d}{c+d}\Bigr)^{c+d}\\ &\Bigr(1+\frac{a}{c}\Bigr)^c\cdot \Bigr(1+\frac{b}{d}\Bigr)^d\leq \Bigr(\frac{a+b+c+d}{c+d}\Bigr)^{c+d}\\ &\Bigr(\frac{c+d}{a+b+c+d}\Bigr)^{c+d}\cdot \Bigr(1+\frac{a}{c}\Bigr)^c\cdot \Bigr(1+\frac{b}{d}\Bigr)^d\leq 1\\ &(a+c)^c\cdot (b+d)^d\cdot (c+d)^{c+d}\leq c^c\cdot d^d (a+b+c+d)^{c+d}. \end{align}

### Acknowledgment

This problem by Dan Sitaru from the Romanian Mathematical Magazine has been kindly posted at the CutTheKnotMath facebook page by Kevin Soto Palacios (Peru), along with his solution. Solution 2 is by Amit Tagi; Solution 3 is by Dan Sitaru.