# lowest common multiple inequality

I've been stuck trying to explicitly prove the following two inequality concerning the lowest common multiple:

$${\frac { \left( \min \left( x,y \right) \right) ^{z}}{z!}}\lt (lcm(x,y))^{z}\leq\frac{1}{2}\, \left( \max \left( x,y \right) \right) ^{z}({x}^{z}+{y}^{z}) \quad \, \forall x,y,z \in \mathbb N \,\backslash {\{1}\}$$

Which is supported by the truth value of the inequality:

$$\frac{1}{2} \left( {x}^{z}+{y}^{z} \right) \left( \max \left( x,y \right) \right) ^{z}- \left( lcm(x,y) \right) ^{z}-{\frac { \left( \min \left( x,y \right) \right) ^{z}}{z!}}+\frac{1}{2} \geq 0$$

$$\forall x,y,z \in \mathbb N$$

The first inequality is clear, since $lcm(x,y)\geq \min(x,y)$ and $z! > 1$.
For the second, notice that both $x \max(x,y)$ and $y \max(x,y)$ are $\geq lcm(x,y)$ since, $lcm(x,y)\leq xy$, so
$$\frac{1}{2}\, \left( \max \left( x,y \right) \right) ^{z}({x}^{z}+{y}^{z}) \\ = \frac{1}{2} \left( ({x\max \left( x,y \right)})^{z}+({y\max \left( x,y \right)})^{z} \right) \\ \geq \frac{1}{2} (lcm(x, y)^z + lcm(x,y)^z) \\ = lcm(x,y)^z$$