Combinatorial Inequalities

Tereza solved the following inequality for $x\in\mathbb{N}$: $${x+1\choose5}\geq5\cdot{x-1\choose 5}$$

(1) First, Tereza converted binomial coefficients into expressions with factorials: $$\frac{(x+1)!}{5!\cdot(x-4)!}\geq5\cdot\frac{(x-1)!}{5!\cdot(x-6)!}$$

(2) She expanded factorials: $$\frac{(x+1)x(x-1)(x-2)(x-3)(x-4)!}{5!\cdot(x-4)!}\geq5\cdot\frac{(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)!}{5!\cdot(x-6)!}$$ (3) She simplified expressions on both sides of the inequality by cancelling common factorials in the numerators and denominators: $$\frac{(x+1)x(x-1)(x-2)(x-3)}{5!}\geq5\cdot\frac{(x-1)(x-2)(x-3)(x-4)(x-5)}{5!}$$ (4) Next, she multiplied both sides of the inequality by $5!$: $$(x+1)x(x-1)(x-2)(x-3)\geq5\cdot(x-1)(x-2)(x-3)(x-4)(x-5)$$ (5) She then divided both sides of the inequality by the expression $(x-1)(x-2)(x-3)$: $$(x+1)x\geq5(x-4)(x-5)$$ (6) Tereza expanded the parentheses and simplified the resulting quadratic inequality to its standard form: \begin{aligned} x^2+x&\geq5(x^2-4x-5x+20)\cr x^2+x&\geq5x^2-45x+100\cr 4x^2-46x+100&\leq0\cr 2x^2-23x+50&\leq0 \end{aligned} (7) She found the roots of the corresponding quadratic equation: $$2x^2-23x+50=0$$ $$x_1\approx2.91,\quad x_2\approx 8.59$$ (8) Finally, Tereza determined the solution of the inequality, considering only the set of natural numbers: $$K=\left\{3,4,5,6,7,8\right\}$$ Tereza’s classmates commented on her solution. Which of them commented incorrectly?

• Adam says that Tereza’s solution is correct.
• David believes that Tereza made a mistake in step (5). He argues that dividing the inequality by the expression $(x-1)(x-2)(x-3)$, could lead to lost solutions. The problem is that one cannot divide an inequality by an expression without knowing its sign.
• Peter agrees with David. He claims that after accounting for the lost solutions, the correct result should be: $$K=\left\{1,2,3,4,5,6,7,8\right\}$$
• Nikola thinks that Tereza forgot to use the conditions for the existence of binomial coefficients. For a binomial coefficient ${n\choose k}$, it holds that $n\geq k\geq 0$. Thus, for the existence of the binomial coefficients $${x+1\choose5}\quad\mbox{and}\quad{x-1\choose 5}$$ it must hold that: $$(x+1\geq5\wedge x-1\geq5)\Rightarrow(x\geq4\wedge x\geq6)\Rightarrow x\geq6$$
• Sandra adds that if we had known the condition $x\geq6$ from the beginning, we could have divided by the expression $(x-1)(x-2)(x-3)$ in step (5), since we would have known that this expression is positive.
• Tom says that all indications suggest the final solution must be $K=\left\{6,7,8\right\}$.