Hello I have been blasting at this inequality proof and it is just not doing what I want it to do:

Prove that $2^{n-1}(a_1a_2\ldots a_n + 1) \geq (1+a_1)(1+a_2)\ldots(1+a_n)$ assuming that $a_1,a_2,\dots, a_n \geq 1$.$\:\:\:$

So the base case is pretty trivial, which leads me to assume that this inequality holds for $k$, so I want to prove $P(k+1)$. I do the following:

$$ (1+a_1)(1+a_2)\ldots(1+a_k)(1+a_{k+1})\leq 2^{k-1}(a_1a_2\ldots a_k + 1)(a_{k+1} + 1) $$ I tried methods like setting the second "$1$" to $a_{k+1}$ and then I have $2^k(a_1\ldots a_k + 1)a_{k+1}$ but I can't assume from there that $2^k(a_1...a_ka_{k+1} +a_{k+1})$ and the second $a_{k+1}$ becomes $1$. So maybe my method is not good from the get-go. Thank you!

  • 1
    $\begingroup$ Do the $n=2$ case separately, and then use that in the induction step. $\endgroup$
    – Calvin Lin
    Mar 12, 2014 at 22:00
  • $\begingroup$ Not very helpful, but you can rewrite the inequality to a more symmetric form: $$\frac{1 + a_1 a_2 \dots a_n}{2} \geq \left(\frac{1 + a_1}{2}\right) \left(\frac{1 + a_2}{2}\right) \dots \left(\frac{1 + a_n}{2}\right).$$ In other words, the function $f(x) = \frac{1 + x}{2}$ satisfies $f(x) f(y) \leq f(xy)$ for $x,y \geq 1$. $\endgroup$
    – TMM
    Mar 12, 2014 at 22:06
  • $\begingroup$ @CalvinLin I don't really understand how that will help, so then I know that $(1+a_1)(1+a_2) \leq 2(a_1a_2 + 1)$ (which is easy to show why it is true) how to use it in the inductive step, do you mean for me to do something like $(1+a_1)(1+a_2)\ldots(1+a_k)(1+a_{k+1}) \leq 2(a_1a_2 + 1)(a_3 + 1)\ldots(a_{k+1} + 1)$? $\endgroup$
    – InsigMath
    Mar 12, 2014 at 22:10

2 Answers 2


The inequality is equivalent to: $$ \left(\frac{1+a_1}{2}\right)\cdots \left(\frac{1+a_n}{2}\right) \le \left(\frac{1+a_1\cdots a_n}{2}\right) $$ Assuming the induction hypothesis we want to show: $$ \left(\frac{1+a_1}{2}\right)\cdots \left(\frac{1+a_{n+1}}{2}\right) \le \left(\frac{1+a_1\cdots a_{n+1}}{2}\right) $$ It is enough to show: $$ \left(\frac{1+a_1\cdots a_n}{2}\right)\cdot \left(\frac{1+a_{n+1}}{2}\right) \le \left(\frac{1+a_1\cdots a_{n+1}}{2}\right) $$ Or $$ 1+a_1\cdots a_n+a_{n+1}+a_1\cdots a_{n+1} \le 2(1+a_1\cdots a_{n+1}) $$ enough to show $$ a_1\cdots a_n + a_{n+1} \le 1+a_1\cdots a_{n+1} $$ Or $$ a_{n+1}-1\le (a_{n+1}-1)a_1\cdots a_n $$ or just $1\le a_1\cdots a_n$.

  • $\begingroup$ Oh wow that is a pretty cool proof! Thank you @Tao, for a split second though I thought you were terence tao :P ... anyway thank you very much! $\endgroup$
    – InsigMath
    Mar 12, 2014 at 22:13
  • $\begingroup$ you're welcome! $\endgroup$
    – TTY
    Mar 12, 2014 at 22:22

I followed your induction strategy. Starting from: $$(1+a_1)\dots(1+a_{k+1})\le 2^{k-1}(a_1\dots a_k+1)(a_{k+1}+1)$$ Let's take $A = a_1 \dots a_k$. $$2^{k-1}(a_1\dots a_k+1)(a_{k+1}+1)=2^{k}(A.a_{k+1} + 1)+2^{k-1}(A-A.a_{k+1}-1+a_{k+1})$$ The first part of the right formula is what we want. We have to prove that the second part is negative to prove $P(k+1)$.

That part can be rewritten $2^{k-1}(A-A.a_{k+1}-1+a_{k+1})=2^{k-1}(A-1)(1-a_{k+1})\le 0$ because $a_{k+1} \ge 1$ and $A \ge 1$. So finally, $$(1+a_1)\dots(1+a_{k+1})\le 2^{k}(a_1\dots a_{k+1}+1)$$


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