For real numbers $a,b,c$ we have $a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}\{(a-b)^2+(b-c)^2+(c-a)^2\}\geqq0$ where equality holds $a=b=c$. In using this, $\left(\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\right)^2=\left(\frac{xy}{z}\right)^2+\left(\frac{yz}{x}\right)^2+\left(\frac{zx}{y}\right)^2+2(x^2+y^2+z^2)$ $\geqq\frac{xy}{z}\cdot\frac{yz}{x}+\frac{yz}{x}\cdot\frac{zx}{y}+\frac{zx}{y}\cdot\frac{xy}{z}+2(x^2+y^2+z^2)=3(x^2+y^2+z^2)=75$. Since $x>0,y>0,z>0$, we obtain $\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\geqq\sqrt{75}$.The equality holds $x^2+y^2+z^2=25$ and $x=y=z\ (x>0,y>0,z>0)\Longleftrightarrow x=y=z=\frac{5}{\sqrt{3}}$. The desired minimum value is $5\sqrt{3}$.

incorrect.

Solution with am-gm $S = \frac{xy}{z} + \frac{xy}{z} + \frac{xy}{z}$ $\frac{S}{3} \ge (xyz)^(%Error. "cbrt" is a bad command. {3})$. From the given statement, we can get that $\frac{25}{3} = \frac{x^2 + y^2 + z^2}{3} \ge ((xyz)^2)^(%Error. "cbrt" is a bad command. {3})$. To minimize $S/3$, we maximize the Rhs of the first inequality. From the 2nd inequality, we see that the maximum of that is $\frac{5}{\sqrt{3}}$, so the minimum of $S$ is $5\sqrt{3}$.

You guys need to be more careful with your inequality signs. In any event, as kunny did, we get: $Q^2\geq \displaystyle \sum \frac {x^2y^2}{z^2} + 2(x^2+y^2+z^2)\geq 3(x^2+y^2+z^2)$ by rearrangement. So $Q\geq \sqrt {75}$, equality holds iff $x=y=z$

Can't we just use Lagrange multipliers to trivialize this problem?

Lagrange Multipliers are not the 1000-pound gorilla of inequality tools. In this case, the desired lagrange multiplier equations are pretty ugly.

(x²y²+y²z²+z²x²)²=(A+B+C)²>=3(AB+BC+CA)=3(xyz)²(x²+y²+z²)=75(xyz)² so sum(xy/z)>=sqrt(75) the equality holds if and only if x=y=z=sqrt(25/3) bye![/code]

Nice solution! math_sipo. math_sipo wrote: $(x^2y^2+y^2z^2+z^2x^2)^2=(A+B+C)^2 \geqq 3(AB+BC+CA)$ $=3(xyz)^2(x^2+y^2+z^2)=75(xyz)^2$ so $\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y} \geqq \sqrt{75}$ the equality holds if and only if $x=y=z=\sqrt{\frac{25}{3}}$ bye![/code] edited by kunny

If we replace $x,y,z$ with $\frac{5}{\sqrt 3}+x_1,\frac{5}{\sqrt 3}+y_1,\frac{5}{\sqrt 3}+z_1$, we get by rearrangment that $\frac{xy}{z}+\frac{xz}{y}+{yz}{x} \geq x+y+z = 5 \sqrt 3 +x_1+y_1+z_1$ Therefore it is at least $ 5 \sqrt 3$ and this can be achieved with $x_1+y_1+z_1=0$ And $x=y=z$