# A Problem from Croatian TST

Leo Giugiuc has posted a problem from a Croatian TST (*Team Selection Test*) at the CutTheKnotMath facebook page. Solution 1 below is by Dan Sitaru and Leo Giugiuc.

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Copyright © 1996-2018 Alexander BogomolnyFor positive $x,y,z,\;$ find the minimum of $\displaystyle\frac{x^2+y^2+z^2}{xy+yz}.$

### Solution 1

Assume, WLOG, that $x^2+y^2+z^2=1.\;$ Then we can specify $t,\phi\in\left( 0,\frac{\pi}{2}\right)\;$ such that $x=\sin t\cos\phi,\;$ $y=\cos t,\;$ $z=\sin t\sin\phi.\;$ The task is to find $\displaystyle\min\left(\frac{1}{\sin t\cos t\,(\sin\phi+\cos\phi )}\right).$

But $\displaystyle \sin t\cos t=\frac{1}{2}\sin 2t\le \frac{1}{2},\;$ with equality for $\displaystyle t=\frac{\pi}{4}.\;$

Also, $\displaystyle\sin\phi+\cos\phi=\sqrt{2}\sin\left(\phi+\frac{\pi}{4}\right)\le \sqrt{2},\;$ with equality for $\displaystyle \phi=\frac{\pi}{4}.\;$

Hence, the required minimum is $\displaystyle \sqrt{2},\;$ which is achieved for $\displaystyle x=z=\frac{1}{2}k\;$ and $\displaystyle y=\frac{1}{\sqrt{2}}k,\;$ $k\gt 0.$

### Solution 2

With the same assumption $x^2+y^2+z^2=1\;$ as in the first solution, the task is to find $\max(xy+yz).\;$ We shall consider the expression $xy+yz\;$ as the scalar product of two-dimensional vectors: $xy+yz=(x,z)\cdot (y,y).$

$xy+yz=(x,z)\cdot (y,y)=\sqrt{x^2+z^2}\sqrt{2y^2}\cos\alpha,\;$ where $\alpha\;$ is the angle between the two vectors. We have

$\displaystyle\begin{align} \sqrt{x^2+z^2}\sqrt{2y^2}\cos\alpha &=\sqrt{1-y^2}\sqrt{2y^2}\cos\alpha\\ &\le\sqrt{1-y^2}\sqrt{2y^2}\\ &=\sqrt{2(1-y^2)y^2}\\ &\le \sqrt{2\left(1-\frac{1}{2}\right)\frac{1}{2}}\\ &= \sqrt{\frac{1}{2}}. \end{align}$

So that the minimum of the original experssion is $\sqrt{2}.\;$ That minimum is achieved for $\alpha=0,\;$ i.e., when $x=z\;$ and $\displaystyle y^2=\frac{1}{2},\;$ meaning that $\displaystyle x=z=\frac{1}{2}\;$ and $\displaystyle y=\frac{1}{\sqrt{2}},\;$ if subject to $x^2+y^2+z^2=1.\;$ There is an infinity of solutions proportional to $\displaystyle\left( \frac{1}{2},\frac{1}{2},\frac{1}{\sqrt{2}}\right).$

(I placed additional solutions on a separate page.)

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Copyright © 1996-2018 Alexander Bogomolny63422237 |