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Suppose $x_0, \cdots, x_n$ are positive reals. Suppose that: $$\sum_{i = 0}^n \frac{1}{1+x_i} \leq 1$$ Then show that: $$\prod_{i=0}^{n} x_i \geq n^{n+1} $$

I got to this problem by rewriting problem $665$ in Andreescu's book. This feels like something pretty standard, but I can't figure it out.

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  • $\begingroup$ It feels GM-HM-ish, but I'm terrible at inequalities and haven't figured out how to get to the finish line yet. $\endgroup$
    – Brian Tung
    Commented Oct 8, 2019 at 20:46
  • $\begingroup$ The assumption is that the harmonic mean of the $x_i+1$ is at least $n+1$. Is it true this implies the harmonic mean of $x_i$ is at least $n$? If so, the result follows from GM-HM. Alternatively maybe we should apply GM-HM immediately to get $\prod_{i=0}^n (x_i+1) \ge (n+1)^{n+1}$, and somehow work from there. $\endgroup$
    – vujazzman
    Commented Oct 8, 2019 at 21:15
  • $\begingroup$ Inequality doesn't seem right. Let n=1 and $x_0=1$ and $x_1=2$. $1 \times 2$ is not greater than $1^2$. $\endgroup$
    – Tahir Imanov
    Commented Oct 8, 2019 at 21:16
  • $\begingroup$ No, it should be right now. For $n=1$, certainly. $\endgroup$
    – mtheorylord
    Commented Oct 8, 2019 at 21:21
  • $\begingroup$ I will give a few hints and if needed present the solution later; first notice that by making $x_0$ smaller we can get equality in the original inequality while lhs of the requested result decreases, so it is enough to prove it then; change variables to $y_k=\frac{1}{x_k+1}$ so $0<y_k<1, \sum {y_k}=1$ and rewrite the inequality in terms of $y_k$. Then using that $1-y_k=\sum_{m\ne k}{y_m}$ apply the mean inequality there and multiply the resulting $n+1$ inequalities to get the result $\endgroup$
    – Conrad
    Commented Oct 8, 2019 at 21:30

2 Answers 2

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Let $y_i:=\dfrac{1}{1+x_i}$ for $i=0,1,2,\ldots,n$. Then, $y_0,y_1,y_2,\ldots,y_n$ are positive real numbers such that $\sum\limits_{i=0}^n\,y_i\leq 1$. We want to prove that $$\prod_{i=0}^n\,\left(\frac{1-y_i}{y_i}\right)\geq {n^{n+1}}.$$

Let $[n]:=\{0,1,2,\ldots,n\}$. To prove the last inequality, we note that $$1-y_i\geq \sum_{j\in[n]\setminus\{i\}}\,y_j\geq n\,\left(\prod_{j\in[n]\setminus\{i\}}\,y_j\right)^{\frac{1}{n}}$$ for each $i\in[n]$. Thus, $$\prod_{i=0}^n\,\left(\frac{1-y_i}{y_i}\right)\geq \prod_{i=0}^n\,\left(\frac{n\,\left(\prod\limits_{j\in[n]\setminus\{i\}}\,y_j\right)^{\frac{1}{n}}}{y_i}\right)={n^{n+1}}\,.$$ The equality holds if and only if $y_0=y_1=y_2=\ldots=y_n=\dfrac1{n+1}$.

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  • $\begingroup$ Thanks, what lead you to consider $1-y_i$? $\endgroup$
    – mtheorylord
    Commented Oct 8, 2019 at 21:23
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    $\begingroup$ Not sure what you mean but I tried solving the problem by assuming that $\sum\limits_{i=0}^n\,y_i=1$ first, and then $1-y_i$ is simply the sum $\sum\limits_{j\in[n]\setminus\{i\}}\,y_j$. This is a homogenization trick that shows up in many inequality problems. $\endgroup$
    – Batominovski
    Commented Oct 8, 2019 at 21:26
  • $\begingroup$ Alright, I'll make sure to remember it. $\endgroup$
    – mtheorylord
    Commented Oct 8, 2019 at 21:28
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Because by AM-GM $$\prod_{i=0}^n\frac{x_i}{1+x_i}\geq\prod_{i=0}^n\sum_{k\neq i}\frac{1}{1+x_k}\geq\prod_{i=0}^n\left(n\prod_{k\neq i}\frac{1}{\sqrt[n]{1+x_k}}\right)=\frac{n^{n+1}}{\prod\limits_{i=0}^n(1+x_i)}$$ and we are done!

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