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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?

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

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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 $xyz>0$.

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