inequality

**lllll** · October 21st 2009, 10:30 PM

I'm having a great deal of difficulty trying to show that if $x, \ y, \ z \ \in \mathbb{R}^{+}$ and that $x \leq y+z$ then $\frac{x}{x+1} \leq \frac{y}{y+1} + \frac{z}{z+1}$

I've tryed working this thing out backwards, taking $\frac{x}{x+1} \leq \frac{y}{y+1} + \frac{z}{z+1}$ and trying to get $x \leq y+z$ but I got stuck in a loop or had a residue of 1.

I also tryed multiplying $x \leq y+z$ by their respective denominators, and their common denominator, but that let nowhere. Squaring didn't help either.

If anybody has any idea of how to approach this, it would be much appreciated.

**BobP** · October 22nd 2009, 12:57 AM

There's probably something cleverer and easier but this seems to work.

Add the fractions on the RHS, cross multiply, expand the brackets and cancel down.
I think that you finish up with $x<xyz+2yz+y+z,$ which is clearly true.
The steps are reversable, that is you can start with $x<y+z$ and say that since $x,y$ and $z$ are positive then $x<xyz+2yz+y+z,$ and so on.

**tonio** · October 22nd 2009, 04:46 AM

Originally Posted by lllll

I'm having a great deal of difficulty trying to show that if $x, \ y, \ z \ \in \mathbb{R}^{+}$ and that $x \leq y+z$ then $\frac{x}{x+1} \leq \frac{y}{y+1} + \frac{z}{z+1}$

I've tryed working this thing out backwards, taking $\frac{x}{x+1} \leq \frac{y}{y+1} + \frac{z}{z+1}$ and trying to get $x \leq y+z$ but I got stuck in a loop or had a residue of 1.

I also tryed multiplying $x \leq y+z$ by their respective denominators, and their common denominator, but that let nowhere

$\color{red}\mbox{Well, it led me to the solution: I did this and got that }x < y+z+xyz+2yz$ $\color{red}\mbox{ and since we're given }x < y+z\,\mbox{ then we're done!}$

$\color{blue}Tonio$

Squaring didn't help either.

If anybody has any idea of how to approach this, it would be much appreciated.

.

**Amer** · October 22nd 2009, 06:49 AM

Originally Posted by lllll

I'm having a great deal of difficulty trying to show that if $x, \ y, \ z \ \in \mathbb{R}^{+}$ and that $x \leq y+z$ then $\frac{x}{x+1} \leq \frac{y}{y+1} + \frac{z}{z+1}$

I've tryed working this thing out backwards, taking $\frac{x}{x+1} \leq \frac{y}{y+1} + \frac{z}{z+1}$ and trying to get $x \leq y+z$ but I got stuck in a loop or had a residue of 1.

I also tryed multiplying $x \leq y+z$ by their respective denominators, and their common denominator, but that let nowhere. Squaring didn't help either.

If anybody has any idea of how to approach this, it would be much appreciated.

since x,y,z all positive integers and

$x\leq y+z$ so

$x > y , x > z$

$x+1 > y+1 ....... x+1 > z+1$

$\frac{1}{x+1 } < \frac{1}{y+1} ,.....and....., \frac{1}{x+1} < \frac{1}{z+1}$

$\frac{x}{x+1} \leq \frac{y}{x+1} + \frac{z}{x+1}$ but

$\frac{y}{x+1} < \frac{y}{y+1}..... and ....\frac{z}{x+1} < \frac{z}{z+1}$

so

$\frac{x}{x+1} \leq \frac{y}{x+1} + \frac{z}{x+1} \leq \frac{y}{y+1} + \frac{z}{z+1}$

**proscientia** · October 22nd 2009, 07:19 AM

Originally Posted by Amer

$x\leq y+z$ so

$x > y , x > z$

Sorry, that’s not true. $1\leqslant2+3$ but $1\not>2,\ 1\not>3.$

One way to prove the inequality is as follows.

Note that for all $t\in\mathbb R^+,$ the function $f(t)=\frac t{t+1}$ is strictly increasing. Hence $x\leqslant y+z$ $\implies$ $\frac x{x+1}\leqslant\frac{y+z}{y+z+1}.$ It therefore suffices to show that $\frac{y+z}{y+z+1}\leqslant\frac y{y+1}+\frac z{z+1}.$

Suppose to the contrary that $\frac{y+z}{y+z+1}>\frac y{y+1}+\frac z{z+1}.$ Then, by cross-multiplying the denominators and patiently expanding and simplifying, I think you should get the get the contradiction $0>yz(y+z+2).$

I admit that this may not be the most elegant solution, but at least it works.

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