Trouble Understanding Proof Using Cauchy's Inequality

[moe darklight]

I'm having trouble understanding a proof in a book I'm reading. It's not really homework because I'm reading it on my own time, but it seems more appropriate to post here than the general math forum.

The exercise is to show that:

$$x+y+z \leq 2\left\{\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\right\}$$

The book says that to prove it, "We apply Cauchy's inequality to the splitting:

$$x+y+z=\frac{x}{\sqrt{y+z}}\sqrt{y+z}+\frac{y}{\sqrt{x+z}}\sqrt{x+z}+\frac{z}{\sqrt{x+y}}\sqrt{x+y}$$"

First, I've never seen set brackets {} used like this. Is it the same as normal brackets in this context (multiply what's inside the brackets by 2)?

And I really don't understand how to prove the first statement with the second... am I supposed to apply Cauchy's inequality to every element of the right hand of the second equation?

If I do that, after simplifying, I get:

$$x+y+z\leq\sqrt{\frac{x^2}{y+z}}\sqrt{y+z}+\sqrt{\frac{y^2}{x+z}}\sqrt{x+z}+\sqrt{\frac{z^2}{x+y}}\sqrt{x+y}$$

Which looks like nothing to me... it just goes back to x+y+z=x+y+z, which doesn't help me much.

I'm really stuck here please help. Thanks!

 

[rasmhop]

Curly brackets mean exactly the same as ordinary parentheses in this context (in a few contexts (, [ and { can all be defined to take a special meaning, but ordinarily they are all different ways of typesetting).

Cauchy's inequality is of the form
$$(ad+be+cf)^2 \leq (a^2+b^2+c^2)(d^2+e^2+f^2)$$
so with your splitting you get:
$$\begin{align*} (x+y+z)^2 &= \left(\frac{x}{\sqrt{y+z}}\sqrt{y+z}+\frac{y}{\sqrt{x+z}}\sqrt{x+z}+\frac{z}{\sqrt{x+y}}\sqrt{x+y}\right)^2 \\ &\leq \left(\left(\frac{x}{\sqrt{y+z}\right)^2+\left(\frac{y}{\sqrt{x+z}}\right)^2+\left(\frac{z}{\sqrt{x+y}}\right)^2}\right)\left(\sqrt{y+z}^2+\sqrt{x+z}^2+\sqrt{x+y}^2\right) \\ &= \left(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\right)\left(2(x+y+z)\right) \end{align*}$$
Now cancel one factor of x+y+z and you have your solution (I assume it was implicit that the x,y,z were positive).

 

[moe darklight]

Oh ok, thanks :)