Solve the chain inequality (x + 1)(x + 2) >= (x + 3)(x + 4) >= (x + 5)(x + 6).

Step-by-step solution

1. 1

   Split the chain into two separate inequalities. A statement $A\ge B\ge C$ is exactly the pair $A\ge B$ and $B\ge C$; the final answer is the intersection of the two solution sets. Trying to handle all three at once is what makes this problem look harder than it is.

2. 2

   Expand the first inequality and watch $x^2$ cancel.

   $x^2+3x+2\ \ge\ x^2+7x+12\ \Longrightarrow\ 3x+2\ge 7x+12\ \Longrightarrow\ -10\ge 4x\ \Longrightarrow\ x\le -\frac{5}{2}$

   Because both sides are monic quadratics, the $x^2$ terms disappear and what looked quadratic is really linear.

3. 3

   Do the same with the second inequality.

   $x^2+7x+12\ \ge\ x^2+11x+30\ \Longrightarrow\ 7x+12\ge 11x+30\ \Longrightarrow\ -18\ge 4x\ \Longrightarrow\ x\le -\frac{9}{2}$

4. 4

   Intersect the two conditions. We need $x\le -\tfrac52$ and $x\le -\tfrac92$. Since $-\tfrac92<-\tfrac52$, the second condition is the binding one:

   $x\le -\frac{9}{2}\qquad\text{i.e.}\qquad x\in\left(-\infty,\ -\tfrac92\right]$

5. 5

   Test the boundary and a point on each side. At $x=-\tfrac92$ the three products are $8.75\ \ge\ 0.75\ \ge\ 0.75$ — true, with equality on the right, exactly as expected at the endpoint. At $x=-5$: $12\ge 2\ge 0$, true. At $x=-4$ (just outside): $6\ge 0\ge 2$ is false, since $0\ge 2$ fails. The boundary is therefore in the right place and the inequality is non-strict there.