# A Problem from Croatian TST, Proof Variants

Leo Giugiuc has posted a problem from a Croatian TST (Team Selection Test) at the CutTheKnotMath facebook page, and a solution with Dan Sitaru. That and one additional solution have been documented on a separate page. The problem has elicited a lively response on social networks. Below, I list several submitted proofs.

### Solution 3 (Alexander Price)

Denote the original expression $\displaystyle E=\frac{x^2+y^2+z^2}{xy+yz}$ and note that $\displaystyle-\frac{2xz}{y(x+z)} \ge -\frac{x+z}{2y}.$ Thus we get

$\displaystyle E \ge \frac{x+z}{2y}+\frac{y}{x+z} \ge 2\cdot\sqrt(1/2)=\sqrt(2).$

### Solution 4 (Kunihiko Chikaya)

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

For equality,

$\displaystyle\frac{\displaystyle\frac{z}{y}+\frac{x}{y}}{2}=\frac{1}{\displaystyle\frac{z}{y}+\frac{x}{y}},$

which gives $\displaystyle\frac{z}{y}-\frac{x}{y}=0,\;$ or $x=y,\;$ and subsequently $y=\sqrt{2}x\;$ such that $f(1,\sqrt{2},1)=\sqrt{2}.$

### Solution 5 (Nassim Nicholas Taleb)

Equating the partial derivatives of $\displaystyle f(x,y,y)=\frac{x^2+y^2+z^2}{xy+yz}$ to $0,\;$ produces three equations:

\displaystyle\begin{align} f_{x}&=\frac{2x}{xy+yz}-\frac{y(x^2+y^2+z^2)}{(xy+yz)^2}=0,\\ f_{x}&=\frac{2y}{xy+yz}-\frac{(x+z)(x^2+y^2+z^2)}{(xy+yz)^2}=0,\\ f_{x}&=\frac{2z}{xy+yz}-\frac{y(x^2+y^2+z^2)}{(xy+yz)^2}=0. \end{align}

The first and the second equations lead to $x=z.\;$ Substituting this into the second equation yields $y=\sqrt{2}z.$ At these values $\displaystyle f=\sqrt{2}.$

### Solution 6 (Imad Zak and Phan Loc Son)

\displaystyle\begin{align} \text{L.H.S.} &=\frac{(x^2+\frac{1}{2}y^2)+(\frac{1}{2}y^2+z^2)}{xy+yz}\\ &=\frac{2xy\sqrt{\frac{1}{2}}+2zy\sqrt{\frac{1}{2}}}{xy+yz}\\ &=\sqrt{2}\frac{1}{2}\frac{xy+yz}{xy+yz}\\ &=\sqrt{2} \end{align}

by AM-GM, and equality holds for $x=z=y/\sqrt{2}.$

### Solution 7 (Kunihiko Chikaya)

Observe that

$\displaystyle x^2+y^2+z^2-\sqrt{2}(z+x)y=\left(y-\frac{z+x}{\sqrt{2}}\right)^2+\frac{1}{2}(z-x)^2\ge 0.$

The equality is achieved for $z=x\;$ and $\displaystyle y=\frac{z+x}{\sqrt{2}}.\;$ Hence, $f_{min}=d(k,\sqrt{2}k,k)=\sqrt{2}.$

### A Sample of Optimization Problems III

• Mathematicians Like to Optimize
• Mathematics in Pizzeria
• The Distance to Look Your Best
• Building a Bridge
• Linear Programming
• Residence at an Optimal Distance
• Distance Between Projections
• Huygens' Problem
• Optimization in a Crooked Trapezoid
• Greatest Difference in Arithmetic Progression
• Area Optimization in Trapezoid
• Minimum under Two Constraints
• Optimization with Many Variables
• Minimum of a Cyclic Sum with Logarithms
• A Problem with a Magical Solution from Secrets in Inequalities
• Leo Giugiuc's Optimization with Constraint
• Problem 4033 from Crux Mathematicorum
• An Unusual Problem by Leo Giugiuc
• A Cyclic Inequality With Constraint in Two Triples of Variables
• Two Problems by Kunihiko Chikaya
• An Inequality and Its Modifications