Inequality with two variables and a conditional equation

Question

Can someone help me solve this inequality?

Let $x$ and $y$ be real numbers with $(x + 1) (y + 2) = 8$. Prove: $$(xy - 10)^2 \geq 64$$ Furthermore, determine all pairs $(x,y)$ of real numbers for which equality holds.

Answer 1

Here's a quick algebraic proof: Note that clearly $y \neq -2$ (otherwise the product would be 0). Thus we get $x = \frac{8}{y+2} - 1 = \frac{6-y}{y+2}$. Now we have $xy = \frac{y(6-y)}{y+2}$ and some quick sketching and checking extrema points should give the inequality. However we can plug in and prove it directly: the condition to prove becomes

$$(xy-10)^2 \geq 8^2 \iff (xy-18)(xy-2) \geq 0 \iff \left( \frac{y(6-y)}{y+2} - 18\right)\left(\frac{y(6-y)}{y+2} - 2\right) \geq 0$$

The expression can be rewritten as $$\frac{1}{(y+2)^2}(-y^2 + 6y - 18y - 36)(-y^2 +6y - 2y - 4) \geq 0$$ and now this immediately becomes

$$\frac{1}{(y+2)^2}(y+6)^2(y-2)^2 \geq 0,$$

which is clearly true. $\square$

We have equality when $y=-6$ or $2$, so $(x,y) = (-3,-6)$ or $(1,2)$.