problem:
Let $x_{1}, x_{2}, \ldots, x_{n}$ be positive real numbers, and let

$$
S=x_{1}+x_{2}+\cdots+x_{n}
$$

Prove that

$$
\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1+x_{n}\right) \leq 1+S+\frac{S^{2}}{2!}+\frac{S^{3}}{3!}+\cdots+\frac{S^{n}}{n!}
$$

solution:
Let $\sigma_{k}$ be the $k$ th symmetric polynomial, namely

$$
\sigma_{k}=\sum_{\substack{|S|=k \\ S \subseteq\{1,2, \ldots, n\}}} \prod_{i \in S} x_{i},
$$

and more explicitly

$$
\sigma_{1}=S, \quad \sigma_{2}=x_{1} x_{2}+x_{1} x_{3}+\cdots+x_{n-1} x_{n}, \quad \text { and so on. }
$$

Then

$$
\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1+x_{n}\right)=1+\sigma_{1}+\sigma_{2}+\cdots+\sigma_{n}
$$

The expansion of

$$
S^{k}=\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{k}=\underbrace{\left(x_{1}+x_{2}+\cdots+x_{n}\right)\left(x_{1}+x_{2}+\cdots+x_{n}\right) \cdots\left(x_{1}+x_{2}+\cdots+x_{n}\right)}_{k \text { times }}
$$

has at least $k$ ! occurrences of $\prod_{i \in S} x_{i}$ for each subset $S$ with $k$ indices from $\{1,2, \ldots, n\}$. In fact, if $\pi$ is a permutation of $S$, we can choose each $x_{\pi(i)}$ from the $i$ th factor of $\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{k}$. Then each term appears at least $k$ ! times, and

$$
S^{k} \geq k!\sigma_{k} \Longleftrightarrow \sigma_{k} \leq \frac{S^{k}}{k!}
$$

Summing the obtained inequalities for $k=1,2, \ldots, n$ yields the result.

problem_match: # Problem 1

solution_match: # Solution 1


solution:
By AM-GM,

$$
\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1+x_{n}\right) \leq\left(\frac{\left(1+x_{1}\right)+\left(1+x_{2}\right)+\cdots+\left(1+x_{n}\right)}{n}\right)^{n}=\left(1+\frac{S}{n}\right)^{n}
$$

By the binomial theorem,

$$
\left(1+\frac{S}{n}\right)^{n}=\sum_{k=0}^{n}\binom{n}{k}\left(\frac{S}{n}\right)^{k}=\sum_{k=0}^{n} \frac{1}{k!} \frac{n(n-1) \ldots(n-k+1)}{n^{k}} S^{k} \leq \sum_{k=0}^{n} \frac{S^{k}}{k!}
$$

and the result follows.
Comment: Maclaurin's inequality states that

$$
\frac{\sigma_{1}}{n} \geq \sqrt{\frac{\sigma_{2}}{\binom{n}{2}}} \geq \cdots \geq \sqrt[k]{\frac{\sigma_{k}}{\binom{n}{k}}} \geq \cdots \geq \sqrt[n]{\frac{\sigma_{n}}{\binom{n}{n}}}
$$

Then $\sigma_{k} \leq\binom{ n}{k} \frac{S^{k}}{n^{k}}=\frac{1}{k!} \frac{n(n-1) \ldots(n-k+1)}{n^{k}} S^{k} \leq \frac{S^{k}}{k!}$.

solution_match: # Solution 2


problem:
Prove that the equation

$$
6\left(6 a^{2}+3 b^{2}+c^{2}\right)=5 n^{2}
$$

has no solutions in integers except $a=b=c=n=0$.

solution:
We can suppose without loss of generality that $a, b, c, n \geq 0$. Let $(a, b, c, n)$ be a solution with minimum sum $a+b+c+n$. Suppose, for the sake of contradiction, that $a+b+c+n>0$. Since 6 divides $5 n^{2}, n$ is a multiple of 6 . Let $n=6 n_{0}$. Then the equation reduces to

$$
6 a^{2}+3 b^{2}+c^{2}=30 n_{0}^{2}
$$

The number $c$ is a multiple of 3 , so let $c=3 c_{0}$. The equation now reduces to

$$
2 a^{2}+b^{2}+3 c_{0}^{2}=10 n_{0}^{2}
$$

Now look at the equation modulo 8:

$$
b^{2}+3 c_{0}^{2} \equiv 2\left(n_{0}^{2}-a^{2}\right) \quad(\bmod 8)
$$

Integers $b$ and $c_{0}$ have the same parity. Either way, since $x^{2}$ is congruent to 0 or 1 modulo 4 , $b^{2}+3 c_{0}^{2}$ is a multiple of 4 , so $n_{0}^{2}-a^{2}=\left(n_{0}-a\right)\left(n_{0}+a\right)$ is even, and therefore also a multiple of 4 , since $n_{0}-a$ and $n_{0}+a$ have the same parity. Hence $2\left(n_{0}^{2}-a^{2}\right)$ is a multiple of 8 , and

$$
b^{2}+3 c_{0}^{2} \equiv 0 \quad(\bmod 8)
$$

If $b$ and $c_{0}$ are both odd, $b^{2}+3 c_{0}^{2} \equiv 4(\bmod 8)$, which is impossible. Then $b$ and $c_{0}$ are both even. Let $b=2 b_{0}$ and $c_{0}=2 c_{1}$, and we find

$$
a^{2}+2 b_{0}^{2}+6 c_{1}^{2}=5 n_{0}^{2}
$$

Look at the last equation modulo 8:

$$
a^{2}+3 n_{0}^{2} \equiv 2\left(c_{1}^{2}-b_{0}^{2}\right) \quad(\bmod 8)
$$

A similar argument shows that $a$ and $n_{0}$ are both even.
We have proven that $a, b, c, n$ are all even. Then, dividing the original equation by 4 we find

$$
6\left(6(a / 2)^{2}+3(b / 2)^{2}+(c / 2)^{2}\right)=5(n / 2)^{2}
$$

and we find that $(a / 2, b / 2, c / 2, n / 2)$ is a new solution with smaller sum. This is a contradiction, and the only solution is $(a, b, c, n)=(0,0,0,0)$.

problem_match: # Problem 2

solution_match: # Solution


problem:
Let $A_{1}, A_{2}, A_{3}$ be three points in the plane, and for convenience,let $A_{4}=A_{1}, A_{5}=A_{2}$. For $n=1,2$, and 3 , suppose that $B_{n}$ is the midpoint of $A_{n} A_{n+1}$, and suppose that $C_{n}$ is the midpoint of $A_{n} B_{n}$. Suppose that $A_{n} C_{n+1}$ and $B_{n} A_{n+2}$ meet at $D_{n}$, and that $A_{n} B_{n+1}$ and $C_{n} A_{n+2}$ meet at $E_{n}$. Calculate the ratio of the area of triangle $D_{1} D_{2} D_{3}$ to the area of triangle $E_{1} E_{2} E_{3}$.
Answer: $\frac{25}{49}$.

solution:
Let $G$ be the centroid of triangle $A B C$, and also the intersection point of $A_{1} B_{2}, A_{2} B_{3}$, and $A_{3} B_{1}$ 。
![](https://cdn.mathpix.com/cropped/2024_11_22_831b46bbbb7a3a492868g-3.jpg?height=441&width=449&top_left_y=750&top_left_x=769)

By Menelao's theorem on triangle $B_{1} A_{2} A_{3}$ and line $A_{1} D_{1} C_{2}$,

$$
\frac{A_{1} B_{1}}{A_{1} A_{2}} \cdot \frac{D_{1} A_{3}}{D_{1} B_{1}} \cdot \frac{C_{2} A_{2}}{C_{2} A_{3}}=1 \Longleftrightarrow \frac{D_{1} A_{3}}{D_{1} B_{1}}=2 \cdot 3=6 \Longleftrightarrow \frac{D_{1} B_{1}}{A_{3} B_{1}}=\frac{1}{7}
$$

Since $A_{3} G=\frac{2}{3} A_{3} B_{1}$, if $A_{3} B_{1}=21 t$ then $G A_{3}=14 t, D_{1} B_{1}=\frac{21 t}{7}=3 t, A_{3} D_{1}=18 t$, and $G D_{1}=A_{3} D_{1}-A_{3} G=18 t-14 t=4 t$, and

$$
\frac{G D_{1}}{G A_{3}}=\frac{4}{14}=\frac{2}{7}
$$

Similar results hold for the other medians, therefore $D_{1} D_{2} D_{3}$ and $A_{1} A_{2} A_{3}$ are homothetic with center $G$ and ratio $-\frac{2}{7}$.
By Menelao's theorem on triangle $A_{1} A_{2} B_{2}$ and line $C_{1} E_{1} A_{3}$,

$$
\frac{C_{1} A_{1}}{C_{1} A_{2}} \cdot \frac{E_{1} B_{2}}{E_{1} A_{1}} \cdot \frac{A_{3} A_{2}}{A_{3} B_{2}}=1 \Longleftrightarrow \frac{E_{1} B_{2}}{E_{1} A_{1}}=3 \cdot \frac{1}{2}=\frac{3}{2} \Longleftrightarrow \frac{A_{1} E_{1}}{A_{1} B_{2}}=\frac{2}{5}
$$

If $A_{1} B_{2}=15 u$, then $A_{1} G=\frac{2}{3} \cdot 15 u=10 u$ and $G E_{1}=A_{1} G-A_{1} E_{1}=10 u-\frac{2}{5} \cdot 15 u=4 u$, and

$$
\frac{G E_{1}}{G A_{1}}=\frac{4}{10}=\frac{2}{5}
$$

Similar results hold for the other medians, therefore $E_{1} E_{2} E_{3}$ and $A_{1} A_{2} A_{3}$ are homothetic with center $G$ and ratio $\frac{2}{5}$.
Then $D_{1} D_{2} D_{3}$ and $E_{1} E_{2} E_{3}$ are homothetic with center $G$ and ratio $-\frac{2}{7}: \frac{2}{5}=-\frac{5}{7}$, and the ratio of their area is $\left(\frac{5}{7}\right)^{2}=\frac{25}{49}$.

problem_match: # Problem 3

solution_match: Solution


problem:
Let $S$ be a set consisting of $m$ pairs $(a, b)$ of positive integers with the property that $1 \leq a<$ $b \leq n$. Show that there are at least

$$
4 m \frac{\left(m-\frac{n^{2}}{4}\right)}{3 n}
$$

triples $(a, b, c)$ such that $(a, b),(a, c)$, and $(b, c)$ belong to $S$.

solution:
Call a triple $(a, b, c)$ good if and only if $(a, b),(a, c)$, and $(b, c)$ all belong to $S$. For $i$ in $\{1,2, \ldots, n\}$, let $d_{i}$ be the number of pairs in $S$ that contain $i$, and let $D_{i}$ be the set of numbers paired with $i$ in $S$ (so $\left|D_{i}\right|=d_{i}$ ). Consider a pair $(i, j) \in S$. Our goal is to estimate the number of integers $k$ such that any permutation of $\{i, j, k\}$ is good, that is, $\left|D_{i} \cap D_{j}\right|$. Note that $i \notin D_{i}$ and $j \notin D_{j}$, so $i, j \notin D_{i} \cap D_{j}$; thus any $k \in D_{i} \cap D_{j}$ is different from both $i$ and $j$, and $\{i, j, k\}$ has three elements as required. Now, since $D_{i} \cup D_{j} \subseteq\{1,2, \ldots, n\}$,

$$
\left|D_{i} \cap D_{j}\right|=\left|D_{i}\right|+\left|D_{j}\right|-\left|D_{i} \cup D_{j}\right| \leq d_{i}+d_{j}-n
$$

Summing all the results, and having in mind that each good triple is counted three times (one for each two of the three numbers), the number of good triples $T$ is at least

$$
T \geq \frac{1}{3} \sum_{(i, j) \in S}\left(d_{i}+d_{j}-n\right)
$$

Each term $d_{i}$ appears each time $i$ is in a pair from $S$, that is, $d_{i}$ times; there are $m$ pairs in $S$, so $n$ is subtracted $m$ times. By the Cauchy-Schwartz inequality

$$
T \geq \frac{1}{3}\left(\sum_{i=1}^{n} d_{i}^{2}-m n\right) \geq \frac{1}{3}\left(\frac{\left(\sum_{i=1}^{n} d_{i}\right)^{2}}{n}-m n\right) .
$$

Finally, the sum $\sum_{i=1}^{n} d_{i}$ is $2 m$, since $d_{i}$ counts the number of pairs containing $i$, and each pair $(i, j)$ is counted twice: once in $d_{i}$ and once in $d_{j}$. Therefore

$$
T \geq \frac{1}{3}\left(\frac{(2 m)^{2}}{n}-m n\right)=4 m \frac{\left(m-\frac{n^{2}}{4}\right)}{3 n} .
$$

Comment: This is a celebrated graph theory fact named Goodman's bound, after A. M. Goodman's method published in 1959. The generalized version of the problem is still studied to this day.

problem_match: # Problem 4


problem:
Determine all functions $f$ from the reals to the reals for which
(1) $f(x)$ is strictly increasing,
(2) $f(x)+g(x)=2 x$ for all real $x$, where $g(x)$ is the composition inverse function to $f(x)$.
(Note: $f$ and $g$ are said to be composition inverses if $f(g(x))=x$ and $g(f(x))=x$ for all real x.)

Answer: $f(x)=x+c, c \in \mathbb{R}$ constant.

solution:
Denote by $f_{n}$ the $n$th iterate of $f$, that is, $f_{n}(x)=\underbrace{f(f(\ldots f}_{n \text { times }}(x)))$.
Plug $x \rightarrow f_{n+1}(x)$ in (2): since $g\left(f_{n+1}(x)\right)=g\left(f\left(f_{n}(x)\right)\right)=f_{n}(x)$,

$$
f_{n+2}(x)+f_{n}(x)=2 f_{n+1}(x)
$$

that is,

$$
f_{n+2}(x)-f_{n+1}(x)=f_{n+1}(x)-f_{n}(x)
$$

Therefore $f_{n}(x)-f_{n-1}(x)$ does not depend on $n$, and is equal to $f(x)-x$. Summing the corresponding results for smaller values of $n$ we find

$$
f_{n}(x)-x=n(f(x)-x) .
$$

Since $g$ has the same properties as $f$,

$$
g_{n}(x)-x=n(g(x)-x)=-n(f(x)-x) .
$$

Finally, $g$ is also increasing, because since $f$ is increasing $g(x)>g(y) \Longrightarrow f(g(x))>$ $f(g(y)) \Longrightarrow x>y$. An induction proves that $f_{n}$ and $g_{n}$ are also increasing functions.
Let $x>y$ be real numbers. Since $f_{n}$ and $g_{n}$ are increasing,

$$
x+n(f(x)-x)>y+n(f(y)-y) \Longleftrightarrow n[(f(x)-x)-(f(y)-y)]>y-x
$$

and

$$
x-n(f(x)-x)>y-n(f(y)-y) \Longleftrightarrow n[(f(x)-x)-(f(y)-y)]<x-y
$$

Summing it up,

$$
|n[(f(x)-x)-(f(y)-y)]|<x-y \quad \text { for all } n \in \mathbb{Z}_{>0}
$$

Suppose that $a=f(x)-x$ and $b=f(y)-y$ are distinct. Then, for all positive integers $n$,

$$
|n(a-b)|<x-y
$$

which is false for a sufficiently large $n$. Hence $a=b$, and $f(x)-x$ is a constant $c$ for all $x \in \mathbb{R}$, that is, $f(x)=x+c$.
It is immediate that $f(x)=x+c$ satisfies the problem, as $g(x)=x-c$.

problem_match: # Problem 5