Subfield: Algebra


Question: There are $n \geqslant 3$ positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$. For each $1 \leqslant i \leqslant n$ we let $b_{i}=\frac{a_{i-1}+a_{i+1}}{a_{i}}$ (here we define $a_{0}$ to be $a_{n}$ and $a_{n+1}$ to be $a_{1}$ ). Assume that for all $i$ and $j$ in the range 1 to $n$, we have $a_{i} \leqslant a_{j}$ if and only if $b_{i} \leqslant b_{j}$.

Prove that $a_{1}=a_{2}=\cdots=a_{n}$.

Solution:
Suppose that not all $a_{i}$ are equal. Consider an index $i$ such that $a_{i}$ is maximal and $a_{i+1}<a_{i}$. Then

$$
b_{i}=\frac{a_{i-1}+a_{i+1}}{a_{i}}<\frac{2 a_{i}}{a_{i}}=2
$$

But since $a_{i}$ is maximal, $b_{i}$ is also maximal, so we must have $b_{j}<2$ for all $j \in\{1,2, \ldots, n\}$.

However, consider the product $b_{1} b_{2} \ldots b_{n}$. We have

$$
\begin{aligned}
b_{1} b_{2} \ldots b_{n} & =\frac{a_{n}+a_{2}}{a_{1}} \cdot \frac{a_{1}+a_{3}}{a_{2}} \cdot \ldots \cdot \frac{a_{n-1}+a_{1}}{a_{n}} \\
& \geqslant 2^{n} \frac{\sqrt{a_{n} a_{2}} \sqrt{a_{1} a_{3}} \ldots \sqrt{a_{n-1} a_{1}}}{a_{1} a_{2} \ldots a_{n}} \\
& =2^{n},
\end{aligned}
$$

where we used the inequality $x+y \geqslant 2 \sqrt{x y}$ for $x=a_{i-1}, y=a_{i+1}$ for all $i \in\{1,2, \ldots, n\}$ in the second row.

Since the product of all $b_{i}$ is at least $2^{n}$, at least one of them must be greater than 2 , which is a contradiction with the previous conclusion.

Thus, all $a_{i}$ must be equal.
Taking $a_{i}$ such that it is maximal among $a_{1}, \ldots, a_{n}$, we obtain $b_{i} \leqslant 2$. Thus $b_{i} \leqslant 2$ for all $j \in\{1,2, \ldots, n\}$.

Consider the product $b_{1} b_{2} \ldots b_{n}$. We have

$$
\begin{aligned}
b_{1} b_{2} \ldots b_{n} & =\frac{a_{n}+a_{2}}{a_{1}} \cdot \frac{a_{1}+a_{3}}{a_{2}} \cdot \ldots \cdot \frac{a_{n-1}+a_{1}}{a_{n}} \\
& \geqslant 2^{n} \frac{\sqrt{a_{n} a_{2}} \sqrt{a_{1} a_{3}} \ldots \sqrt{a_{n-1} a_{1}}}{a_{1} a_{2} \ldots a_{n}} \\
& =2^{n},
\end{aligned}
$$

where we used the inequality $x+y \geqslant 2 \sqrt{x y}$ for $x=a_{i-1}, y=a_{i+1}$ for all $i \in\{1,2, \ldots, n\}$ in the second row.

The equation above then gives $2^{n} \geqslant b_{1} \cdots b_{n} \geqslant 2^{n}$, which together with $b_{i} \leqslant 2$ for all $j \in\{1,2, \ldots, n\}$ implies that $b_{j}=2$ for all $j \in\{1,2, \ldots, n\}$. Since we have $b_{1}=$ $b_{2}=\cdots=b_{n}$, the condition that $a_{i} \leqslant a_{j} \Longleftrightarrow b_{i} \leqslant b_{j}$ gives that $a_{1}=a_{2}=\cdots=a_{n}$.
Taking $a_{i}$ such that it is maximal among $a_{1}, \ldots, a_{n}$, we obtain $b_{i} \leqslant 2$. Thus $b_{i} \leqslant 2$ for all $j \in\{1,2, \ldots, n\}$.

Then

$$
\begin{aligned}
2 n & \geqslant b_{1}+\cdots+b_{n}=\frac{a_{n}}{a_{1}}+\frac{a_{2}}{a_{1}}+\frac{a_{1}}{a_{2}}+\frac{a_{3}}{a_{2}}+\cdots+\frac{a_{n-1}}{a_{n}}+\frac{a_{1}}{a_{n}} \\
& \geqslant 2 n \sqrt[n]{\frac{a_{n}}{a_{1}} \cdot \frac{a_{2}}{a_{1}} \cdot \frac{a_{1}}{a_{2}} \cdot \frac{a_{3}}{a_{2}} \cdots \frac{a_{n-1}}{a_{n}} \cdot \frac{a_{1}}{a_{n}}}=2 n \cdot 1=2 n,
\end{aligned}
$$

where we used the AM-GM inequality.

It follows that all $b_{j}$ 's are equal which as in Solution 2 gives $a_{1}=a_{2}=\cdots=a_{n}$.
By assumption $a_{i} b_{i}=a_{i-1}+a_{i+1}$ for $i=\{1,2, \ldots, n\}$, hence,

$$
\sum_{i=1}^{n} a_{i} b_{i}=2 \sum_{i=1}^{n} a_{i}
$$

Since $a_{i} \leqslant a_{j}$ if and only if $b_{i} \leqslant b_{j}$, the Chebyshev's inequality implies

$$
\left(\sum_{i=1}^{n} a_{i}\right) \cdot\left(\sum_{i=1}^{n} b_{i}\right) \leqslant n \cdot \sum_{i=1}^{n} a_{i} b_{i}=2 n \cdot \sum_{i=1}^{n} a_{i}
$$

and so $\sum_{i=1}^{n} b_{i} \leqslant 2 n$. On the other hand, we have

$$
\sum_{i=1}^{n} b_{i}=\sum_{i=1}^{n} \frac{a_{i-1}}{a_{i}}+\sum_{i=1}^{n} \frac{a_{i+1}}{a_{i}}=\sum_{i=1}^{n} \frac{a_{i-1}}{a_{i}}+\sum_{i=1}^{n} \frac{a_{i}}{a_{i-1}}=\sum_{i=1}^{n}\left(\frac{a_{i-1}}{a_{i}}+\frac{a_{i}}{a_{i-1}}\right)
$$

so we can use the AM-GM inequality to estimate

$$
\sum_{i=1}^{n} b_{i} \geqslant \sum_{i=1}^{n} 2 \sqrt{\frac{a_{i-1}}{a_{i}} \cdot \frac{a_{i}}{a_{i-1}}}=2 n
$$

We conclude that we must have equalities in all the above, which implies $\frac{a_{i-1}}{a_{i}}=\frac{a_{i}}{a_{i-1}}$ and consequently $a_{i}=a_{i-1}$ for all positive integers $i$. Hence, all $a$ 's are equal.
Consider the product $b_{1} b_{2} \ldots b_{n}$. We have

$$
\begin{aligned}
b_{1} b_{2} \ldots b_{n} & =\frac{a_{n}+a_{2}}{a_{1}} \cdot \frac{a_{1}+a_{3}}{a_{2}} \cdot \ldots \cdot \frac{a_{n-1}+a_{1}}{a_{n}} \\
& \geqslant 2^{n} \frac{\sqrt{a_{n} a_{2}} \sqrt{a_{1} a_{3}} \ldots \sqrt{a_{n-1} a_{1}}}{a_{1} a_{2} \ldots a_{n}} \\
& =2^{n},
\end{aligned}
$$

where we used the inequality $x+y \geqslant 2 \sqrt{x y}$ for $x=a_{i-1}, y=a_{i+1}$ for all $i \in\{1,2, \ldots, n\}$ in the second row.

By assumption $a_{i} b_{i}=a_{i-1}+a_{i+1}$ for $i=\{1,2, \ldots, n\}$, hence,

$$
\sum_{i=1}^{n} a_{i} b_{i}=2 \sum_{i=1}^{n} a_{i}
$$

Since $a_{i} \leqslant a_{j}$ if and only if $b_{i} \leqslant b_{j}$, the Chebyshev's inequality implies

$$
\left(\sum_{i=1}^{n} a_{i}\right) \cdot\left(\sum_{i=1}^{n} b_{i}\right) \leqslant n \cdot \sum_{i=1}^{n} a_{i} b_{i}=2 n \cdot \sum_{i=1}^{n} a_{i}
$$

and so $\sum_{i=1}^{n} b_{i} \leqslant 2 n$. 

As above we show that $\sum_{i=1}^{n} b_{i} \leqslant 2 n$ and $\prod_{i=1}^{n} b_{i} \geqslant 2^{n}$. We now use the AM-GM inequality and the first inequality to get

$$
\prod_{i=1}^{n} b_{i} \leqslant\left(\frac{1}{n} \sum_{i=1}^{n} b_{i}\right)^{n} \leqslant\left(\frac{1}{n} \cdot 2 n\right)^{n}=2^{n}
$$

This implies that we must have equalities in all the above. In particular, we have equality in the AM-GM inequality, so all $b$ 's are equal. Since we have $b_{1}=$ $b_{2}=\cdots=b_{n}$, the condition that $a_{i} \leqslant a_{j} \Longleftrightarrow b_{i} \leqslant b_{j}$ gives that $a_{1}=a_{2}=\cdots=a_{n}$.
Let $a_{i}$ to be minimal and $a_{j}$ maximal among all $a$ 's. Then

$$
b_{j}=\frac{a_{j-1}+a_{j+1}}{a_{j}} \leqslant \frac{2 a_{j}}{a_{j}}=2=\frac{2 a_{i}}{a_{i}} \leqslant \frac{a_{i-1}+a_{i+1}}{a_{i}}=b_{i}
$$

and by assumption $b_{i} \leqslant b_{j}$. Hence, we have equalities in the above so $b_{j}=2$ so $a_{j-1}+$ $a_{j+1}=2 a_{j}$ and therefore $a_{j-1}=a_{j}=a_{j+1}$. We have thus shown that the two neighbors of a maximal $a$ are also maximal. By an inductive argument all $a$ 's are maximal, hence equal.
Choose an arbitrary index $i$ and assume without loss of generality that $a_{i} \leqslant a_{i+1}$. (If the opposite inequality holds, reverse all the inequalities below.) By induction we will show that for each $k \in \mathbb{N}_{0}$ the following two inequalities hold

$$
\begin{gathered}
a_{i+1+k} \geqslant a_{i-k}
\end{gathered}
\tag{1}
$$
$$
\begin{gathered}
a_{i+1+k} a_{i+1-k} \geqslant a_{i-k} a_{i+k}
\end{gathered}
\tag{2}
$$



(where all indices are cyclic modulo $n$ ). Both inequalities trivially hold for $k=0$.

Assume now that both inequalities hold for some $k \geqslant 0$. The inequality $a_{i+1+k} \geqslant a_{i-k}$ implies $b_{i+1+k} \geqslant b_{i-k}$, so

$$
\frac{a_{i+k}+a_{i+2+k}}{a_{i+1+k}} \geqslant \frac{a_{i-1-k}+a_{i+1-k}}{a_{i-k}}
$$

We may rearrange this inequality by making $a_{i+2+k}$ the subject so

$$
a_{i+2+k} \geqslant \frac{a_{i+1+k} a_{i-1-k}}{a_{i-k}}+\frac{a_{i+1-k} a_{i+1+k}-a_{i+k} a_{i-k}}{a_{i-k}} \geqslant \frac{a_{i+1+k} a_{i-1-k}}{a_{i-k}},
$$

where the last inequality holds by (2). It follows that

$$
a_{(i+1)+(k+1)} a_{(i+1)-(k+1)} \geqslant a_{i+(k+1)} a_{i-(k+1)}
$$

i.e. the inequality (2) holds also for $k+1$. Using (1) we now get

$$
a_{(i+1)+(k+1)} \geqslant \frac{a_{i+k+1}}{a_{i-k}} a_{i-(k+1)} \geqslant a_{i-(k+1)},
$$

i.e. (1) holds for $k+1$.

Now we use the inequality (1) for $k=n-1$. We get $a_{i} \geqslant a_{i+1}$, and since at the beginning we assumed $a_{i} \leqslant a_{i+1}$, we get that any two consecutive $a$ 's are equal, so all of them are equal.
We first prove the following claim by induction:

Claim 1: If $a_{k} a_{k+2}<a_{k+1}^{2}$ and $a_{k}<a_{k+1}$, then $a_{j} a_{j+2}<a_{j+1}^{2}$ and $a_{j}<a_{j+1}$ for all $j$.

We assume that $a_{i} a_{i+2}<a_{i+1}^{2}$ and $a_{i}<a_{i+1}$, and then show that $a_{i-1} a_{i+1}<a_{i}^{2}$ and $a_{i-1}<a_{i}$.

Since $a_{i} \leq a_{i+1}$ we have that $b_{i} \leq b_{i+1}$. By plugging in the definition of $b_{i}$ and $b_{i+1}$ we have that

$$
a_{i+1} a_{i-1}+a_{i+1}^{2} \leq a_{i}^{2}+a_{i+2} a_{i} \tag{3}
$$

Using $a_{i} a_{i+2}<a_{i+1}^{2}$ we get that

$$
a_{i+1} a_{i-1}<a_{i}^{2} \tag{4}
$$

Since $a_{i}<a_{i+1}$ we have that $a_{i-1}<a_{i}$, which concludes the induction step and hence proves the claim.

We cannot have that $a_{j}<a_{j+1}$ for all indices $j$. Similar as in the above claim, one can prove that if $a_{k} a_{k+2}<a_{k+1}^{2}$ and $a_{k+2}<a_{k+1}$, then $a_{j+1}<a_{j}$ for all $j$, which also cannot be the case. Thus we have that $a_{k} a_{k+2} \geq a_{k+1}^{2}$ for all indices $k$.

Next observe (e.g. by taking the product over all indices) that this implies $a_{k} a_{k+2}=a_{k+1}^{2}$ for all indices $k$, which is equivalent to $b_{k}=b_{k+1}$ for all $k$ and hence $a_{k+1}=a_{k}$ for all $k$.
Define $c_{i}:=\frac{a_{i}}{a_{i+1}}$, then $b_{i}=c_{i-1}+1 / c_{i}$. Assume that not all $c_{i}$ are equal to 1. Since, $\prod_{i=1}^{n} c_{i}=1$ there exists a $k$ such that $c_{k} \geqslant 1$. From the condition given in the problem statement for $(i, j)=(k, k+1)$ we have

$$
c_{k} \geqslant 1 \Longleftrightarrow c_{k-1}+\frac{1}{c_{k}} \geqslant c_{k}+\frac{1}{c_{k+1}} \Longleftrightarrow c_{k-1} c_{k} c_{k+1}+c_{k+1} \geqslant c_{k}^{2} c_{k+1}+c_{k}
\tag{5}
$$

Now since $c_{k+1} \leqslant c_{k}^{2} c_{k+1}$, it follows that

$$
c_{i-1} c_{i} c_{i+1} \geqslant c_{i} \Longrightarrow\left(c_{i-1} \geqslant 1 \text { or } c_{i+1} \geqslant 1\right)
\tag{6}
$$

So there exist a set of at least 2 consecutive integers, such that the corresponding $c_{i}$ are greater or equal to one. By the innitial assumption there must exist an index $\ell$, such that $c_{\ell-1}, c_{\ell} \geqslant 1$ and $c_{\ell+1}<1$. We distinguish two cases:

Case 1: $c_{\ell}>c_{\ell-1} \geqslant 1$

From $c_{\ell-1} c_{\ell} c_{\ell+1}<c_{\ell}^{2} c_{\ell+1}$ and the inequality (5), we get that $c_{\ell+1}>c_{\ell} \geqslant 1$, which is a contradiction to our choice of $\ell$.

Case 2: $c_{\ell-1} \geqslant c_{\ell} \geqslant 1$

Once again looking at the inequality (5) we can find that

$$
c_{\ell-2} c_{\ell-1} c_{\ell} \geqslant c_{\ell-1}^{2} c_{\ell} \Longrightarrow c_{\ell-2} \geqslant c_{\ell-1} \tag{7}
$$

Note that we only needed $c_{\ell-1} \geqslant c_{\ell} \geqslant 1$ to show $c_{\ell-2} \geqslant c_{\ell-1} \geqslant 1$. So using induction we can easily show $c_{\ell-s-1} \geqslant c_{\ell-s}$ for all $s$.

So

$$
c_{1} \leqslant c_{2} \leqslant \cdots \leqslant c_{n} \leqslant c_{1} \tag{8}
$$

a contradiction to our innitial assumption.

So our innitial assumtion must have been wrong, which implies that all the $a_{i}$ must have been equal from the start.