# Algebraic-Geometric Inequality

### Solution 1

In complex numbers, Let $u=e^{i\pi/6},\,$ $v=e^{i\pi/4},\,$ and $w=e^{5i\pi/12}.\,$ We have $\sqrt{x^2-\sqrt{3}xy+y^2}=|x-yu|,\,$ $\sqrt{y^2-\sqrt{2}yz+z^2}=|y-zv|=|u||y-zv|=|yu-zw|.\,$ It follows that

\displaystyle\begin{align} \sqrt{x^2-\sqrt{3}xy+y^2} &+ \sqrt{y^2-\sqrt{2}yz+z^2} = |x-yu|+|yu-zw|\\ &\ge |x-yu+yu-zw|=|x-zw|\\ &=\sqrt{x^2-xz\left(\frac{\sqrt{6}-\sqrt{2}}{2}\right)+z^2}\\ &\ge\sqrt{x^2-xz+z^2}. \end{align}

### Solution 2

Consider triangles $ABC\,$ and $ACD\,$ such that $AB=x,\,$ $AC=y,\,$ $AD=z,\,$ $\angle BAC=\pi/6,\,$ $\angle CAD=\pi/4.\,$ Then $BC=\sqrt{x^2-\sqrt{3}xy+y^2}\,$ and $CD=\sqrt{y^2-\sqrt{2}yz+z^2}.\,$ Also, $\angle BAD=75^{\circ},\,$ $BD=\sqrt{z^2-zx\cos 75^{\circ}+x^2}.$

Since $\cos 75^{\circ}\lt\cos 60^{\circ}=1/2,\,$ $z^2-zx\cos 75^{\circ}+x^2\gt z^2-zx+x^2.\,$ Now,

$BC+CD\ge BD\gt \sqrt{z^2-zx+x^2}$

which proves the required inequality.

### Solution 3

\displaystyle\begin{align} \sqrt{x^2-\sqrt{3}xy+y^2} &+ \sqrt{y^2-\sqrt{2}yz+z^2}\\ = &\sqrt{\left(\frac{\sqrt{3}}{2}x-y\right)^2+\left(\frac{x}{2}\right)^2}+\sqrt{\left(y-\frac{z}{\sqrt{2}}\right)^2+\left(\frac{z}{\sqrt{2}}\right)^2}\\ &\ge\sqrt{\left(\frac{\sqrt{3}}{2}x-\frac{z}{\sqrt{2}}\right)^2+\left(\frac{x}{2}+\frac{z}{\sqrt{2}}\right)^2}\\ &=\sqrt{x^2+z^2-\frac{\sqrt{3}-1}{\sqrt{2}}zx}\\ &\gt\sqrt{z^2-xz+x^2}, \end{align}

since $\sqrt{2}+1\gt\sqrt{3}.$

This completes the proof.

### Acknowledgment

Dan Sitaru has kindly posted the above problem (from his book "Math Accent") at the CutTheKnotMath facebook page. Solution 1 is by Leo Giugiuc; Solution 2 is by Ravi Prakash and, independently, by Chris Kyriazis; Solution 3 is by Nguyen Minh Triet and, indepedently, by Diego Alvariz.