inequality

AndrewTom

Prove that $\frac{x+y}{1+x+y} < \frac{x}{1+x} + \frac{y}{1+y},$ for all positive numbers $x$ and $y$ .

**Posted:** Mon Oct 26, 2009 4:01 am

darkdieuguerre

Proof

**Posted:** Mon Oct 26, 2009 4:25 am

_________________
$10^{12}$ bites do not make a terrabyte.

akashram

Just use andrscu's lemma and you are done

**Posted:** Sat Oct 31, 2009 8:53 pm

randomguy64

what's that?

**Posted:** Sat Oct 31, 2009 9:33 pm

AndrewTom

Titu Andrescu's lemma is known as Titu's lemma. See here: http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=264227

**Posted:** Sun Nov 01, 2009 4:47 am

varunrocks · Riemann Hypothesis Online Joined: 25 Dec 2007 Posts: 292

How do u use T2's Lemma if
you get that
(x/x+1)+(y/y+1)>=(x+y)^2/x+y+2
how does that help?

**Posted:** Sun Nov 01, 2009 9:56 am

Agr_94_Math · Yang-Mills Theory Offline Joined: 17 Feb 2008 Posts: 714

Akashram is right. But I don't know why people call it Titu's lemma or Engel Cauchy or whatever.
For example, using the Holder's Inequality if I say
$\sum \frac{x_i^k}{a_i}\ge \frac{(\sum x_i)^k}{n^{k-2} \sum a_i}$ , will I say it as Agr_94_Math's lemma?

Anyway, as far as the solution is concerned,
writing the $RHS$ as $\sum \frac{x^2}{x+x^2}$ we get that $\sum \frac{x^2}{x+x^2} \ge \frac{(x+y)^2}{x+y+x^2+y^2}$ BY CAUCHY SCHWARZ.
So we have to now prove that $\frac{(x+y)^2}{x+y+x^2+y^2} \ge \frac{x+y}{1+x+y}$ .
Simplifying, it is equivalent to $2xy \ge 0$ which is obviously true and equality and cannot hold as $x,y >0$ .
Thus, we prove the inequality.

Akashram, since you are new, it is no problem but in future see to it that you post your solutions completely when you want to post, so that it avoids spamming.

**Posted:** Thu Nov 05, 2009 7:24 pm