# Inequality using Cauchy Schwarz

I’ve recently been looking into applications of the Cauchy Schwarz inequality.

I’ve seen it stated in the form:

$$(a_1^2 + a_2^2 + … + a_n^2)(b_1^2 + b_2^2 + … + b_n^2) \ge (a_1b_1 + a_2b_2 + … + a_nb_n)^2$$

However I’m struggling to see how it could prove this cyclic sum inequality:

$$\sum_\text{cyc} \frac{x^2}{y + z} \ge \frac{(x+y+z)^2}{2(x+y+z)}$$

Where $$xyz = 1$$ and $$x,y,z$$ $$\epsilon$$ $$\mathbb{R}$$

Any help?

• Is your inequality's right-hand side meant to simplify to $\frac{x+y+z}{2}$?
– J.G.
Dec 4 '20 at 21:26

Assuming $$x, y, z \geq 0$$, the Cauchy-Schwarz inequality gives
\begin{align*} &\left((y + z) + (z + x) + (x + y)\right)\left(\frac{x^2}{y + z} + \frac{y^2}{z + x} + \frac{z^2}{x + y}\right) \\ &\geq \left(\sqrt{y + z}\sqrt{\frac{x^2}{y + z}} + \sqrt{z + x}\sqrt{\frac{y^2}{z + x}} + \sqrt{x + y}\sqrt{\frac{z^2}{x + y}}\right)^2 \\ &= (x + y + z)^2 \end{align*}
and you can divide out $$(y + z) + (z + x) + (x + y) = 2(x + y + z)$$ on both sides to get your inequality.
Write $$a:=y+z,\,b:=z+x,\,c:=x+y$$ (a common strategy with $$3$$-variable cyclic-symmetry problems, especially if $$y+z$$ etc. appears), so the conjectured inequality is $$(a+b+c)\sum_\text{cyc}\frac{x^2}{a}\ge(x+y+z)^2$$. Now take $$a_1=\sqrt{a},\,b_1=\frac{x}{\sqrt{a}}$$ etc.
Note this proof never ever used $$xyz=1$$. Multiplying each of $$x,\,y,\,z$$ by $$\lambda>0$$ also multiplies both sides of your inequality by $$\lambda$$ and preserves its truth, and so whatever proof we obtain should only need to use $$xyz>0$$.