Proof of $xy\ge4$ with $x=z+\frac{1}{p}$ and $y=p+\frac{1}{z}$, where $x,y,z,p \in R$ $x,y,z,p\gt0$

Question

Consider following set of equations: $$x=z+\frac{1}{p}$$ $$y=p+\frac{1}{z}$$ Where $$x,y,z,p \in R$$$$x,y,z,p\gt0$$ How to prove: $$xy\ge4$$

Answer 1

It will be easier if you prove the following first.

Lemma: Let $r>0$. Then $r + \frac{1}{r} \geq 2$.

Proof: Assume toward a contradiction that $r=2-\epsilon$ where $0<\epsilon<2$. Then $r+\frac{1}{r}=2-\epsilon$ and $r^2 + (\epsilon-2)r + 1=0$. Using the quadratic formula and solving for $r$ gives $r = \frac{(2-\epsilon) \pm \sqrt{\epsilon (\epsilon - 4)}}{2}$, where the term under the radical is negative. Thus there is no real solution, which contradicts $r$ being real. $\square$

You can now easily prove $xy\geq 4$, by substituting in for $x$ and $y$ and applying the lemma to $zp$.