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

Question

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"?

Answer 1

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

Answer 2

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 $f''(u) = \frac{2A^2}{(A-u)^3} > 0$.

Answer 3

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.

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

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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.

Answer 4

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.

Answer 5

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.

Answer 6

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.