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.

A Problem from Croatian  TST, Proof Variants: Find minimum of (xx+yy+zz)/(xy+yz)

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,


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, $f_{min}=d(k,\sqrt{2}k,k)=\sqrt{2}.$

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