# An inequality with positive integers

Let $$s$$ be a positive integer. Let $$a_1, a_2,\ldots, a_s$$ be distinct positive integers such that $$a_i\geq2, \ \forall \ i\in\lbrace 1,2,\ldots, s\rbrace$$. Show that $$(a_1a_2\ldots a_s)^2-a_1a_2\ldots a_s+1>\prod\limits_{i=1}^s(a_i^2-a_i+1)-1$$.

I tried to prove it by induction on $$s$$ (for $$s=1$$ and $$s=2$$ the inequality holds; then, without success, I tried to apply the inductive hypothesis, i.e. we know that $$(a_1a_2\ldots a_{s-1})^2+a_1a_2\ldots a_{s-1}+1>\prod\limits_{i=1}^{s-1}(a_i^2-a_i+1)-1$$ and we pass from $$s-1$$ to $$s$$. Any hint is appreciated.

• Are you sure it is not $(a_1a_2\ldots a_s)^2\color{red}{-}a_1a_2\ldots a_s+1$ in the LHS? The way it is now, the statement is too trivial. – Ivan Neretin Apr 15 at 21:42
• You are right. Thanks for this notice. – Alchimist Apr 16 at 7:30

## 2 Answers

There seems to be a typo in the inequality. In its current form, is is trivial (see Servaes' answer). We are going to show a similar and stronger inequality instead.

For any $$n \ge 2$$ and $$a_1, \ldots, a_n \ge 1$$, we have $$\left(\prod_{k=1}^n a_k\right)^2 \color{red}{-} \left(\prod_{k=1}^n a_k\right) + 1 \ge \prod_{k=1}^n (a_k^2 - a_k + 1)$$

Let $$f(x) = x^2 - x + 1$$, above inequality can be rewritten as

$$f\left(\prod_{k=1}^n a_k\right) \ge \prod_{k=1}^n f(a_k) \quad\text{ for } a_1,\ldots,a_n \ge 1$$

For any $$n \ge 2$$, let $$\mathcal{S}_n$$ be the statement that above inequality is true at that paricular $$n$$.

Notice for any $$a, b \ge 1$$,

\begin{align} f(ab)-f(a)f(b) &= (ab)^2 - ab + 1 - (a^2-a+1)(b^2+b+1)\\ &= (a-1)(b-1)(a+b) \ge 0\end{align}

The base statement $$\mathcal{S}_2$$ is true.

Assume $$\mathcal{S}_n$$ is true for a particular $$n$$. For any $$a_1,\ldots,a_{n+1} \ge 1$$, we have

$$f\left(\prod_{k=1}^{n+1} a_k\right) = f\left(a_1\prod_{k=2}^{n+1} a_k\right) \stackrel{\mathcal{S}_2}{\ge} f(a_1)f\left(\prod_{k=2}^{n+1} a_k\right) \stackrel{\mathcal{S}_n, f(a_1) \ge 0}{\ge} f(a_1)\prod_{k=2}^{n+1}f(a_k) = \prod_{k=1}^{n+1}f(a_k)$$ This means $$\mathcal{S}_n \implies \mathcal{S}_{n+1}$$. By induction, $$\mathcal{S}_n$$ is true for all $$n \ge 2$$.

Apply this to the inequality at hand. For any $$a_1,\ldots, a_s \ge 2$$, we have

\begin{align} \left(\prod_{k=1}^s a_k\right)^2 + \left(\prod_{k=1}^s a_k\right) + 1 > &\left(\prod_{k=1}^s a_k\right)^2 - \left(\prod_{k=1}^s a_k\right) + 1 \\ \ge & \prod_{k=1}^s(a_k^2-a_k+1)\\ > & \prod_{k=1}^s(a_k^2-a_k+1) - 1 \end{align}

It seems abundantly clear that $$0 and hence that $$\prod_i(a_i^2-a_i+1)<\prod_ia_i^2=\big(\prod_i a_i\big)^2<(a_1a_2\ldots a_s)^2+a_1a_2\ldots a_s+1.$$