Cauchy-Schwarz inequality and three-letter identities (exercise 1.4 from "The Cauchy-Schwarz Master Class")

Exercise 1.4 from a great book The Cauchy-Schwarz Master Class asks to prove the following:

For all positive $x$, $y$ and $z$, one has $$x+y+z \leq 2 \left(\frac{x^2}{y+z} + \frac{y^2}{x+z} + \frac{z^2}{x+y}\right).$$

Introduction to the exercise says:

There are many situations where Cauchy's inequality conspires with symmetry to provide results that are visually stunning.

How to prove that inequality? And how does one benefit from the "symmetry"? What is the general idea behind this "conspiracy"?

• The back of the book has solutions to the exercises.
– user940
Oct 10, 2011 at 19:09
• (This was my original comment, that I removed by mistake.) I don't think the answers here will explain the conspiracy better than Steele does in his book. (But let's hope that I am wrong about this.) So my recommendation is: read the book! :) Oct 10, 2011 at 19:43
• Uh, I didn't even realize that the book has solutions. I've only printed out the first chapter, but the book is definitely worth buying. Unfortunately, as Martin has noticed, the back of the book doesn't elaborate on the conspiracy. Oct 10, 2011 at 21:10

$(x+y+z)^2 = \left(x\sqrt{\frac{y+z}{y+z}} +y\sqrt{\frac{x+z}{x+z}}+z\sqrt{\frac{x+y}{x+y}}\right)^2\leq\left(\frac{x^2}{y+z} + \frac{y^2}{x+z} + \frac{z^2}{x+y}\right)(y+z+x+z+x+y)$. You will get it. I didn't read that book, but I believe Srivatsan is right.

The inequality is $$f\left(\frac{x+y+z}{3}\right) \leq \frac{f(x)+f(y)+f(z)}{3}$$ for all positive $x,y,z$ with sum $A$, where $f(u) = \frac{u^2}{A-u}$, and $A=x+y+z$.

This is equivalent to $f(u)$ being convex in the interval $(0,A)$. The second derivative is $f''(u) = \frac{2A^2}{(A-u)^3} > 0$.

The inequality $$(x+y+z)^2\le \left(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\right)[(y+z)+(x+z)+(x+y)]$$ can be obtained from Cauchy-Schwarz.

(You use it for triples $\frac{x}{\sqrt{y+z}}$, $\frac{y}{\sqrt{x+z}}$, $\frac{z}{\sqrt{x+y}}$ and $\sqrt{y+z}$, $\sqrt{x+z}$, $\sqrt{x+y}$.)

Now if you cancel $(x+y+z)$ you get the desired inequality.

However, I am not able to explain "the conspiracy", so I left this task for other answerers.

After seeing Byron Schmuland's comment I had a look into the book, and found out that basically the same explanation is given in the back of the book.

There is an inequality which can be deduced by using the Cauchy Schwarz inequality:

Let $a_1,...,a_n$ be real numbers and $b_1,...,b_n>0$. Then $$\sum_{i=1}^n \frac{a_i^2}{b_i} \geq \frac{(a_1+...+a_n)^2}{b_1+...+b_n}$$

The desired inequality is a simple consequence of the above.

• This is called Begstrom inequality. Oct 10, 2011 at 20:38
• Bergström, not Begstrom.
– user940
Jan 8, 2014 at 18:34

I have solved it without using Cauchy's Inequality

We have to prove that

$\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{y+x}=(x+y+z)\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{y+x}-1\right)\geq \left(\dfrac{x+y+z}{2}\right)$

$\implies \dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{y+x}\geq \dfrac{3}{2}$

And this is Nesbitt's Inequality (The proof is given in this link http://en.wikipedia.org/wiki/Nesbitt's_inequality)

Hence Proved.

I think the symmetry here comes from being able to derive a similar set of inequalities incorporating the same splitting method. To build off of @Martin's method, we have $$(x+y+z) \leq 2\left(\frac{x^2}{y+z} + \frac{y^2}{x+z} + \frac{z^2}{x+y}\right)$$ And using the same splitting method we can derive

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

by splitting $$(x+y+z) = \frac{x}{\sqrt{x+z}}\sqrt{x+z} + \frac{y}{\sqrt{x+y}}\sqrt{x+y} + \frac{z}{\sqrt{y+z}}\sqrt{y+z}$$ And similarly $$(x+y+z) \leq 2\left(\frac{x^2}{x+y} + \frac{y^2}{y+z} + \frac{z^2}{x+z}\right)$$

So we have three different upper bounds just by rotating how we decide to perform the splitting.