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Despite many attempts, no one at StackOverflow has succeeded in solving that old question about proving a deceptively simple-looking inequality.

I propose now a weaker and slightly simpler inequality (no more squares) which may perhaps be easier to prove.

Let $x_1,x_2, \ldots ,x_n$ be positive numbers whose product is $1$. Prove or find a counterexample :

$$ \prod_{k=1}^n \big(\frac{1+x_k}{2}\big) \leq \bigg( \frac{x_1+x_2+x_3+ \ldots x_n}{n}\bigg)^{n-1} $$

(this inequality follows from the old one by putting $x_i=a_i^2$ and using Cauchy-Schwarz).

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Ewan Delanoy
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    Using the equal variable method (See http://math.stackexchange.com/questions/51835/a-three-variable-inequality), it suffices to show the case where $x_1 \leq 1 \leq x_2 = \cdots = x_n$. Did you try to do that? –  Jan 17 '13 at 19:08
  • I also think the EV-method works. Take the inverse of this inequality, then Log both sides and apply Corollary 1.6 from Cîrtoaje's article cited above for the function $f(x)=Log(\frac{2}{1+x})$ – ivan Jan 23 '13 at 07:04
  • @ivan : Corollary 1.6 tells us that the extremal value that interests us will be attained at a point where $x_1=x_2=\ldots x_{n-1}<x_n$, but that’s not the end of the story, is it ? What this does is reduce the $n$-variable problem to a two-variables problem. – Ewan Delanoy Jan 23 '13 at 07:19
  • Yes, you are right. It reduces the inequality to a nasty looking two variable problem. – ivan Jan 23 '13 at 12:26

2 Answers2

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I am going to propose something that might work for the reduced 2-variable problem (see the comments above). We need to prove that given $xy^n=1$ then:$$f(x,y)=\left(\frac{x+n y}{n+1}\right)^n-\frac{1+x}{2}\left(\frac{1+y}{2}\right)^n\geq 0$$ Effectively we need to show $f(y^{-n},y)\geq 0$. Now consider the polynomial $y^{n^2}f(y^{-n},y)$. It has a double root at $y=1$.

Moreover it looks like its coefficients can be computed effectively. I suggest it can be shown that there are only 4 sign alternations for this polynomial. Hence using the Descartes' rule of signs the number of positive roots is either 2 (in which case we are done because of the double root at 1) or 4.

If there are 4 roots the remaining 2 roots are both either bigger then 1 or smaller then 1. Hence we are done for at least one of the cases $y>1$ or $y<1$.

This may be enough to establish the general case.

ivan
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As mentioned in comment, one can use equal variable method for $f(x) = \log \left(\frac{1+x}{2}\right)$. Note that $$g(x) = f'(1/x) = \frac{x}{x+1}$$ is strictly concave for $x > 0$. By corollary 1.6, the maximum of LHS is reached when $0 < x_1 \leq 1 \leq x_2 = \cdots = x_n = t$. Substitute $x_1 = \frac{1}{t^{n-1}}$, we have to prove $$\left(\frac{(n-1)t + \frac{1}{t^{n-1}}}{n}\right)^{n-1} \ge \left(\frac{1+t}{2}\right)^{n-1}\left(\frac{1+\frac{1}{t^{n-1}}}{2}\right)$$ for $t \ge 1$. By AMGM applied to $(n-2)$ $(1+t)/2$, and $(1+t)(1+t^{-(n-1)})/4$, we see that $$RHS \leq \left(\frac{(n-2)\left(\frac{1+t}{2}\right) + \frac{1+t+t^{-(n-1)} + t^{-(n-2)}}{4}}{n-1}\right)^{n-1}$$ So it suffices to show that $$\frac{(n-1)t + \frac{1}{t^{n-1}}}{n} \ge \frac{(n-2)\left(\frac{1+t}{2}\right) + \frac{1+t+t^{-(n-1)} + t^{-(n-2)}}{4}}{n-1}$$ Clearing denominator and simplify, $$\Leftrightarrow 4(n-1)^2 t + 4(n-1)t^{-(n-1)} \ge 2n(n-2)t + 2n(n-2) + n (1+t+t^{-(n-1)} + t^{-(n-2)})$$ $$\Leftrightarrow (2n^2-5n+4)t + (3n-4)t^{-(n-1)} \ge 2n^2 - 3n + nt^{-(n-2)} \hspace{5mm} (*)$$ We prove the last inequality now. AM-GM gives $$\frac{n(n-2)}{(n-1)} t^{-(n-1)} + \frac{n}{n-1} \ge nt^{-(n-2)}$$ and $$(2n^2-5n+4)t + \frac{2n^2-5n+4}{n-1}t^{-(n-1)} \ge (2n^2-5n+4)\frac{n}{n-1}$$ (*) follows from adding up the last two inequalities.

  • @EwanDelanoy, I don't understand what you are saying. If we multiply as you said, we get $(n-2) + t^{n-1} \ge (n-1)t$, apparently true by AMGM. In fact, eventually I didn't use $t \ge 1$ in the whole proof. –  Jan 25 '13 at 17:24
  • I was indeed wrong in my former comment (which is why I’ve just deleted it). I’m fully convinced now, answer accepted. – Ewan Delanoy Jan 26 '13 at 09:26
  • @EwanDelanoy, no problem. Unfortunately this method does not apply to the original problem, and I am clueless how to approach it without a tool like EV theorem. –  Jan 26 '13 at 09:44