# Suppose a, b, c are real numbers and x,y,z>=0. Prove the following inequality

1. Apr 8, 2010

### Kizaru

1. The problem statement, all variables and given/known data
Suppose that a, b, c are real numbers and x, y, z >= 0. Prove that

$$\frac{a^2}{x} + \frac{b^2}{y} + \frac{c^2}{z} \geq \frac{ (a+b+c)^2}{x+y+z}$$

2. Relevant equations
Cauchy-Schwarz and Arithmetic Geometric Mean inequalities.

3. The attempt at a solution
I wasn't really sure how to approach this problem. I tried brute forcing a solution by multiplying everything out to get common denominators, but that became a mess. I tried a geometric approach of two vectors but didn't get anywhere with it.

Any help would be appreciated. Thanks.

Last edited: Apr 8, 2010
2. Apr 8, 2010

### lanedance

the tex closing tag is a / rather than backslash for the functions etc.
$$\frac{a^2}{x} + \frac{b^2}{y} + \frac{c^2}{z} \geq \frac{ (a+b+c)^2}{x+y+z}$$

3. Apr 9, 2010

### Staff: Mentor

I'm pretty sure you need x, y, z > 0, otherwise the terms on the left side could be undefined.

You might try a simpler problem, such as
$$\frac{a^2}{x} + \frac{b^2}{y} \geq \frac{ (a+b)^2}{x+y}$$

and see if you can prove that. Doing so might give you some insight on the harder problem.

In any case, proving the original statement is equivalent to proving this statement:
$$\frac{a^2}{x} + \frac{b^2}{y} + \frac{c^2}{z} - \frac{ (a+b+c)^2}{x+y+z}\geq 0$$

I worked on this about a half page or so, but am going to quit for the night.

4. Apr 9, 2010

### Kizaru

Err yes, it should be x,y,z > 0. I haven't touched it since last night, so I'll see where I can get with the simpler problem today.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook