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Reduction formulae

Show that

$$\frac xy + \frac yz + \frac zx \ge 1 + \frac {z + x}{x + y} + \frac {x + y}{z + x}$$

for $x, y, z \gt 0$.

I observed that this is a homogeneous inequality so normalization might work. I tried to set $x = 1$ or $xyz = 1$ or $x + y + z = 1$, but none of these yields a solution. How do I tackle this problem? In particular, what can I do with the term $1$? Any hints will be appreciated.

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Using CS inequality in slightly different way,

$$\frac{x}y + \frac{y}z + \frac{z}x + 1 \geqslant \frac{(x+y+z+x)^2}{xy+yz+zx+x^2} = \frac{(\color{red}{x+y}\:+\:\color{blue}{z+x})^2}{(\color{red}{x+y})\cdot(\color{blue}{z+x})} = \frac{x+y}{z+x}+2 + \frac{z+x}{x+y}$$

The cyclic counterparts of the inequality can be shown similarly.

**Addendum:** *the inequality above is perhaps seen simplest by*

$$\frac{x}y + \frac{y}z + \frac{z}x + 1 = \frac{x^2}{xy}+\frac{y^2}{yz}+\frac{z^2}{zx}+\frac{x^2}{x^2} \geqslant \frac{(x+y+z+x)^2}{xy+yz+zx+x^2} $$

the CS form is called “Titu’s lemma” and is a useful tool to have.

We have

$$\frac xy + \frac yz + \frac zx – 1 – \frac {z + x}{x + y} – \frac {x + y}{z + x} = {\frac { \left( x-y \right) ^{2}}{ \left( x+y \right) y}}+{\frac {x

\left( y-z \right) ^{2}}{z \left( z+x \right) \left( x+y \right) }}+{\frac { \left( x-z \right) ^{2}y}{ \left( x+y \right) xz}}.$$

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