# How to prove specific inequality, assuming $\prod\limits_{i=1}^{n}(a_{i}-1)=1$

Let $n\ge 2$ be postive integer and $a_{i},(i=1,2,\cdots,n)$ be real numbers such that $a_{i}>1,(i=1,2,\cdots,n),\prod_{i=1}^{n}(a_{i}-1)=1$. Show that $$\sum_{i=1}^{n}\dfrac{1}{\displaystyle\sum_{k=1}^{n-1}ka_{i+k-1}}\le\dfrac{1}{n-1}$$ where $a_{n+m}=a_{m},\forall m\ge 1$

It seem use AM-GM inequality. I have try all methods can't solve this problem.

$$\sum_{k=1}^{n-1}ka_{i+k-1}=a_{i}+2a_{i+2}+3a_{i+3}+\cdots+(n-1)a_{i+n-1-1}$$ $$\sum_{i=1}^{n}\dfrac{1}{\displaystyle\sum_{k=1}^{n-1}ka_{i+k-1}}=\sum_{i=1}^{n}\dfrac{1}{a_{i}+2a_{i+2}+3a_{i+3}+\cdots+(n-1)a_{i+n-1-1}}\le\dfrac{1}{n-1}$$

• Wait, for $i=4$ we get that $a_{i+n-1-1} = a_{n+2}$, something doesnt seem right, since we only have $n$ variables? – cirpis Aug 18 '17 at 14:39
• $a_{n+2}=a_{2}$ – function sug Aug 18 '17 at 14:50
• Ah, missed that, thanks! – cirpis Aug 18 '17 at 14:54
• This is work in progress, a quick brainstorming shows that $$\sum_{k=1}^{n−1}ka_{i+k−1}=a_i+2a_{i+2}+3a_{i+3}+...+(n−1)a_{i+n−1−1}>1+2+3+...+n-1=\frac{n(n-1)}{2}$$ or $$\frac{1}{\sum\limits_{k=1}^{n−1}ka_{i+k−1}}<\frac{2}{n(n-1)}$$ and $$\sum\limits_{i=1}^{n}\frac{1}{\sum\limits_{k=1}^{n−1}ka_{i+k−1}}<\frac{2n}{n(n-1)}=\frac{2}{n-1}$$ Question is, how does $\prod\limits_{i=1}^{n}(a_i−1)=1$ improve it? – rtybase Aug 20 '17 at 15:22

## 2 Answers

This problem was proposed by Dmitriy Maximov on XXIV Russian Festival of Math Youth. Here is the official solution.

Denote $a_k^{n-1}=x_k-1$. The product of new positive variables $a_i$ equals 1. Consider the denominator of the first fraction: \begin{align*} x_1+2x_2+3x_3+ \ldots +(n-1)x_{n-1}&=a_1^{n-1}+2a_2^{n-1}+ \ldots (n-1)a_{n-1}^{n-1}+ 1+2+\ldots+(n-1)= \\ =(n-1)+\bigl(a_{n-1}^{n-1}+(n-2)\bigr)&+\bigl(a_{n-1}^{n-1}+a_{n-2}^{n-1}+(n-3)\bigr)+ \bigl(a_{n-1}^{n-1}+a_{n-2}^{n-1}+a_{n-3}^{n-3}+(n-4)\bigr)+\\ \ldots+ & \bigl(a_{n-1}^{n-1}+a_{n-2}^{n-1}+\ldots+a_1^{n-1}\bigr). \end{align*} Now we estimate all summands (except the first) by AM-GM for $n-1$ numbers (some of them equal to 1). This makes denominators lesser, thus fractions greater. The denominator of the first fraction becomes $$(n-1)a_{n-1}+(n-1)a_{n-1}a_{n-2}+(n-1)a_{n-1}a_{n-2}a_{n-3}+\ldots+(n-1)a_{n-1}a_{n-2}\ldots a_{1}+(n-1).$$ Now it suffices to prove that the cyclic sum of the fractions $$\frac{1}{a_{n-1}+a_{n-1}a_{n-2}+\ldots+a_{n-1}a_{n-2}\ldots a_1+1}$$ does not exceed 1. Actually this sum equals 1. For proving this we multiply the second fraction by $a_{n-1}a_{n-2} \ldots a_1$ (I mean, multiply both the numerator and denominator), the third by $a_{n-2}a_{n-3} \ldots a_1$ and so on (the last fraction by $a_1$). The denominators become equal, and the sum of numerators becomes equal to the denominator.

• Nice,Thank you,and I ask it which year,I can't search it,can you give thik links?Thanks – function sug Aug 21 '17 at 12:01
• the problems in Russian may be found on www.guas.info – Fedor Petrov Aug 21 '17 at 12:11

Let $a_i-1=x_i$.

Thus, $\prod\limits_{i=1}^nx_i=1$ and we need to prove that $$\sum_{i=1}^n\frac{1}{\sum\limits_{k=1}^{n-1}ka_{i+k-1}}\leq\frac{1}{n-1}$$ or $$\sum_{cyc}\frac{1}{a_1+2a_2+3a_3+...+(n-1)a_{n-1}}\leq\frac{1}{n-1}$$ or $$\sum_{cyc}\frac{1}{x_1+2x_2+3x_3+...+(n-1)x_{n-1}+1+2+3+...+(n-1)}\leq\frac{1}{n-1}$$ or $$\sum_{cyc}\frac{1}{x_1+2x_2+3x_3+...+(n-1)x_{n-1}+\frac{n(n-1)}{2}}\leq\frac{1}{n-1}$$ or $$\sum_{cyc}\left(\frac{1}{x_1+2x_2+3x_3+...+(n-1)x_{n-1}+\frac{n(n-1)}{2}}-\frac{1}{\frac{n(n-1)}{2}}\right)\leq\frac{1}{n-1}-\frac{2}{n-1}$$ or $$\sum_{cyc}\frac{x_1+2x_2+3x_3+...+(n-1)x_{n-1}}{x_1+2x_2+3x_3+...+(n-1)x_{n-1}+\frac{n(n-1)}{2}}\geq\frac{n}{2}.$$ Now, by C-S $$\sum_{cyc}\frac{x_1+2x_2+3x_3+...+(n-1)x_{n-1}}{x_1+2x_2+3x_3+...+(n-1)x_{n-1}+\frac{n(n-1)}{2}}=$$ $$\sum_{i=1}^n\frac{\sum\limits_{k=1}^{n-1}kx_{i+k-1}}{\sum\limits_{k=1}^{n-1}kx_{i+k-1}+\frac{n(n-1)}{2}}=\sum_{i=1}^n\frac{\left(\sum\limits_{k=1}^{n-1}kx_{i+k-1}\right)^2}{\left(\sum\limits_{k=1}^{n-1}kx_{i+k-1}\right)^2+\frac{n(n-1)}{2}\cdot\sum\limits_{k=1}^{n-1}kx_{i+k-1}}\geq$$ $$\geq\frac{\left(\sum\limits_{i=1}^n\sum\limits_{k=1}^{n-1}kx_{i+k-1}\right)^2}{\sum\limits_{i=1}^n\left(\left(\sum\limits_{k=1}^{n-1}kx_{i+k-1}\right)^2+\frac{n(n-1)}{2}\sum\limits_{k=1}^{n-1}kx_{i+k-1}\right)}=$$ $$=\frac{\frac{n^2(n-1)^2}{4}\sum\limits_{i=1}^nx_i}{\sum\limits_{i=1}^n\left(\left(\sum\limits_{k=1}^{n-1}kx_{i+k-1}\right)^2+\frac{n(n-1)}{2}\sum\limits_{k=1}^{n-1}kx_{i+k-1}\right)}.$$ Thus, it remains to prove that $$\frac{n(n-1)^2}{2}\sum\limits_{i=1}^nx_i\geq\sum\limits_{i=1}^n\left(\left(\sum\limits_{k=1}^{n-1}kx_{i+k-1}\right)^2+\frac{n(n-1)}{2}\sum\limits_{k=1}^{n-1}kx_{i+k-1}\right)$$ or $$\frac{n(n-1)^2}{2}\sum\limits_{i=1}^nx_i\geq\sum\limits_{i=1}^n\left(\sum\limits_{k=1}^{n-1}ka_{i+k-1}\right)^2+\frac{n^2(n-1)^2}{4}\sum\limits_{k=1}^{n}x_k.$$ But $$\sum\limits_{i=1}^n\left(\sum\limits_{k=1}^{n-1}ka_{i+k-1}\right)^2=\sum_{cyc}(x_1+2x_2+3x_3+...+(n-1)x_{n-1})^2=$$ $$=\sum_{cyc}\left(x_1^2+2^2x_2^2+...+(n-1)^2x_{n-1}^2\right)+2\sum_{cyc}\sum_{1\leq j<k\leq n-1}jkx_jx_k=$$ $$=(1^2+2^2+...+(n-1)^2)\sum_{i=1}^nx_i^2+2\sum_{cyc}\sum_{1\leq j<k\leq n-1}jkx_jx_k=$$ $$=\frac{(n-1)n(2n-1)}{6}\sum_{i=1}^nx_i^2+2\sum_{cyc}\sum_{1\leq j<k\leq n-1}jkx_jx_k.$$ Hence, we need to prove that $$\tfrac{n(n-1)^2}{2}\sum_{i=1}^nx_i^2+n(n-1)^2\sum_{1\leq j<k\leq n}x_jx_k\geq$$ $$\geq\tfrac{(n-1)n(2n-1)}{6}\sum_{i=1}^nx_i^2+2\sum_{cyc}\sum_{1\leq j<k\leq n-1}jkx_jx_k+\frac{n^2(n-1)^2}{4}\sum\limits_{k=1}^{n}x_k$$ or $$\left(\tfrac{n(n-1)^2}{2}-\tfrac{(n-1)n(2n-1)}{6}\right)\sum_{i=1}^nx_i^2+n(n-1)^2\sum_{1\leq j<k\leq n}x_jx_k\geq$$ $$\geq2\sum_{cyc}\sum_{1\leq j<k\leq n-1}jkx_jx_k+\frac{n^2(n-1)^2}{4}\sum\limits_{k=1}^{n}x_k$$ or $$\frac{n(n-1)(n-2)}{6}\sum_{i=1}^nx_i^2+n(n-1)^2\sum_{1\leq j<k\leq n}x_jx_k\geq$$ $$\geq2\sum_{cyc}\sum_{1\leq j<k\leq n-1}jkx_jx_k+\frac{n^2(n-1)^2}{4}\sum\limits_{k=1}^{n}x_k.$$ Now, $$\sum\limits_{cyc}\sum\limits_{1\leq j<k\leq n-1}jkx_jx_k=\sum_{cyc}\sum i(i+1)x_ix_{i+1}+\sum_{cyc}\sum i(i+2)x_ix_{i+2}+...=$$ $$=A_1\sum_{cyc}x_1x_2+A_2\sum_{cyc}x_1x_3+...$$ and by Rearrangement $A_1$ is a maximal coefficient between all coefficients $A_k$,

which says that in the left side the coefficient before $\sum\limits_{cyc}x_1x_2$,

which is $n(n-1)^2-2A_1$, will be minimal.

For the rest expressions we can use AM-GM: $$(2A_1-2A_2)\sum_{cyc}x_1x_3+ (2A_1-2A_3)\sum_{cyc}x_1x_4+...\geq K,$$ where the value of $K$ we'll get a bit of later.

Thus, it's enough to prove that $$\frac{n(n-1)(n-2)}{6}\sum_{i=1}^nx_i^2+\left(n(n-1)^2-2A_1\right)\sum_{1\leq j<k\leq n}x_jx_k+K\geq\frac{n^2(n-1)^2}{4}\sum\limits_{k=1}^{n}x_k.$$ Now, $$A_1=1\cdot2+2\cdot3+...+(n-2)(n-1)=\sum_{k=1}^{n-2}k(k+1)=$$ $$=\frac{(n-2)(n-1)(2n-3)}{6}+\frac{(n-2)(n-1)}{2}=\frac{n(n-1)(n-2)}{3}.$$ Thus, it's enough to prove that: $$\tfrac{n(n-1)(n-2)}{6}\sum_{i=1}^nx_i^2+\left(n(n-1)^2-\tfrac{2n(n-1)(n-2)}{3}\right)\sum_{1\leq j<k\leq n}x_jx_k+K\geq\tfrac{n^2(n-1)^2}{4}\sum\limits_{k=1}^{n}x_k$$ or $$\tfrac{n(n-1)(n-2)}{6}\sum_{i=1}^nx_i^2+\tfrac{n(n-1)(n+1)}{3}\sum_{1\leq j<k\leq n}x_jx_k+K\geq\tfrac{n^2(n-1)^2}{4}\sum\limits_{k=1}^{n}x_k.$$ Now, we can get a value of $K$.

Indeed, the equality in the original inequality occurs for $x_1=x_2=...=x_n=1$ and AM-GM saves this thing.

Thus, $$K=\frac{n^2(n-1)^2}{4}\cdot n-\frac{n(n-1)(n-2)}{6}\cdot n-\frac{n(n-1)(n+1)}{3}\cdot\frac{n(n-1)}{2}=$$ $$=\frac{n^2(n-1)(n-2)(n-3)}{12}.$$ Id est, we need to prove that: $$\tfrac{n(n-1)(n-2)}{6}\sum_{i=1}^nx_i^2+\tfrac{n(n-1)(n+1)}{3}\sum_{1\leq j<k\leq n}x_jx_k+\tfrac{n^2(n-1)(n-2)(n-3)}{12}\geq\tfrac{n^2(n-1)^2}{4}\sum\limits_{k=1}^{n}x_k$$ or $$2(n-2)\sum_{i=1}^nx_i^2+4(n+1)\sum_{1\leq j<k\leq n}x_jx_k+n(n-2)(n-3)\geq3n(n-1)\sum\limits_{k=1}^{n}x_k.$$ Now, let $\sum\limits_{i=1}^nx_i=nu$, $\sum\limits_{1\leq i<j\leq n}x_ix_j=\frac{n(n-1)}{2}v^2$, where $v>0$,

$\sum\limits_{1\leq i<j<k\leq n}x_ix_jx_k=\frac{n(n-1)(n-2)}{6}w^3$ and $\prod\limits_{i=1}^nx_i=t^n$, where $t>0$.

Thus, $u\geq v\geq w\geq t$ and we need to prove that $$2(n-2)(n^2u^2-n(n-1)v^2)+4(n+1)\cdot\tfrac{n(n-1)}{2}v^2+n(n-2)(n-3)t^2\geq3n(n-1)\cdot nut$$ or $f(t)\geq0$, where $$f(t)=2(n-2)(nu^2-(n-1)v^2)+2(n^2-1)v^2+(n-2)(n-3)t^2-3n(n-1)ut.$$ But $$f'(t)=2(n-2)(n-3)t-3n(n-1)u\leq u(2n^2-10n+12-3n^2+3n)=u(-n^2-7n+12)\leq0,$$ which says $$f(t)\geq f(v)=$$ $$=2(n-2)(nu^2-(n-1)v^2)+2(n^2-1)v^2+(n-2)(n-3)v^2-3n(n-1)uv=$$ $$=(u-v)(2(n-2)u-(n+1)v),$$ which says that for $$2(n-2)-(n+1)\geq0$$ or $$n\geq5$$ our inequality is proven.

Thus, it remains to prove our inequality for $n\in\{2,3,4\}$.

The proof for $n=4$.

We need to prove that $$8u^2+9v^2+t^2\geq18ut$$ and since $t\leq w$, it's enough to prove that $$8u^2+9v^2+w^2-18uw\geq0.$$ Now, $x_1$, $x_2$, $x_3$ and $x_4$ are positive roots of the equation $$(x-x_1)(x-x_2)(x-x_3)(x-x_4)=0$$ or $$x_4-4ux^3+4v^2x^2-w^3x+t^4=0.$$ Thus, by Rolle the following equation $$(x_4-4ux^3+4v^2x^2-w^3x+t^4)'=0$$ or $$x^3-3ux^2+3v^2-w^3=0$$ has three positive roots.

Thus, there are positives $a$, $b$ and $c$ for which $3u=a+b+c$, $ab+ac+bc=3v^2$ and $abc=w^3$,

which says that it's enough to prove $$8u^2+9v^2+w^2-18uw\geq0$$ for all positives $a$, $b$ and $c$.

But this inequality is a linear inequality of $v^2$, which says that it remains to prove the last inequality for an extremal value (even for a minimal value in our case) of $v^2$, which happens for equality case of two variables.

Since the last inequality is homogeneous, we can assume $b=c=1$ and for $a=t^3$ we need to prove that $$\frac{8(t^3+2)^2}{9}+3(2t^3+1)+t^2-6(t^3+2)t\geq0$$ or $$(t-1)^2(8t^4+16t^3-30t^2+10t+59)\geq0,$$ which is obvious.

The proof for $n=3$.

We need to prove that $$a^2+b^2+c^2+8(ab+ac+bc)-9(a+b+c)\geq0$$ for positives $a$, $b$ and $c$ such that $abc=1$.

But this inequality is a linear inequality of $v^2$ again, which says that it's enough to prove the last inequality for $b=a$ and $c=\frac{1}{a^2}$, which gives $$(a-1)^2(10a^4+2a^3-6a^2+2a+1)\geq0,$$ which is obvious.

For $n=2$ the starting inequality it's just equality.

Done!

• For n=4 and n=3 have other methods to solve it? Thanks – function sug Aug 20 '17 at 23:33
• @function sug I think Mixing Variables method can help, but it's not so nice. – Michael Rozenberg Aug 21 '17 at 0:12
• WOW. That is all I have to say, @MichaelRozenberg. Well done! – bjcolby15 Aug 21 '17 at 1:12
• Beautiful...... – George N. Missailidis Aug 21 '17 at 7:39
• @GeorgeN.Missailidis It gets the job done, and that is a merit in on itself, but I wouldn't call it an especially beautiful proof :D – cirpis Aug 21 '17 at 9:58