Minimizing the sum of fractions

Applying AM-GM inequality , we have:

• $\dfrac{yz}{x}+\dfrac{zx}{y}\geq 2z$

• $\dfrac{zx}{y}+\dfrac{xy}{z}\geq 2x$

• $\dfrac{xy}{z}+\dfrac{yz}{x}\geq 2y$

Therefore, adding these inequalities up gives $\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\geq x+y+z=\boxed{91}$ .

Relevant wiki: Rearrangement Inequality

$\sum_{cyc}\frac{xy}{z}\ge\sum_{cyc} \frac{xy}{y}=x+y+z=91$ rearrangement inequality

Note that $(a-b)^2+(b-c)^2+(c-a)^2 \ge 0$ $\implies a^2+b^2+c^2 \ge ab+bc+ca$ From this we can discern $(a+b+c)^2 \ge 3(ab+bc+ca) \tag{1}$ $3(a^2+b^2+c^2) \ge (a+b+c)^2 \tag{2}$

If $a=\frac{xy}{z}, b=\frac{yz}{x}, c=\frac{zx}{y}$ , then note that from $\text{(1)}$ and $\text{(2)}$

$\left( \sum_{cyc} \frac{xy}{z} \right)^2 \ge 3 \left(\sum_{cyc} x^2 \right) \ge (x+y+z)^2$

From this, we have that the following inequality holds with equality when $x=y=z$ . $\sum_{cyc} \frac{xy}{z} \ge 91$

Set x= y = z