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.

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    $\begingroup$ 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. $\endgroup$ – Ivan Neretin Apr 15 at 21:42
  • $\begingroup$ You are right. Thanks for this notice. $\endgroup$ – Alchimist Apr 16 at 7:30

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


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