Problem 1279

I. solution. If $1 \leq xc$ then $d \geq 5$, on the other hand $a \geq 1$ and thus $b \geq 2$, so

$$\frac{1}{2 a}+\frac{1}{2 b}+\frac{1}{2 c}+\frac{1}{2 d} \leq \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{10}$$

which is indeed less than 1.

If $c<4$, then $a=1, b=2$ and $c=3$, so $d \geq 4$. In this case

$$\frac{1}{[a, b]}+\frac{1}{[b, c]}+\frac{1}{2 c}+\frac{1}{2 d} \leq \frac{1}{2}+\frac{1}{6}+\frac{1}{6}+\frac{1}{8}<1$$

so in this case we have also proved the statement.

Ákos Magyar (Budapest, Fazekas M. Gymn., 1st year)

II. solution. A better estimate can avoid the case analysis. Since $[x, y]$ is a multiple of both $x$ and $y$, $xy$ is a multiple of both $y[x, y]$ and $x[x, y]$. Therefore, the difference of these two latter numbers is also a multiple of $xy$. If this difference is not 0, then its absolute value is at least $xy$. Thus, if $0<x<y$, then

$$y[x, y]-x[x, y] \geq xy$$

from which, after dividing by $[x, y] \cdot x \cdot y$, we get that

$$\frac{1}{x}-\frac{1}{y} \geq \frac{1}{[x, y]}$$

Applying this,

$$\begin{gathered} \frac{1}{[a, b]}+\frac{1}{[b, c]}+\frac{1}{[c, d]}+\frac{1}{[d, e]} \leq \\ \leq\left(\frac{1}{a}-\frac{1}{b}\right)+\left(\frac{1}{b}-\frac{1}{c}\right)+\left(\frac{1}{c}-\frac{1}{d}\right)+\left(\frac{1}{d}-\frac{1}{e}\right)=\frac{1}{a}-\frac{1}{e} \end{gathered}$$

follows, which is indeed less than 1.

Gábor Megyesi (Szeged, Juhász Gy. Teacher Training College 1st year Practice School, 8th grade)

Remarks. 1. Many incorrect solutions were submitted for the problem. Most of them featured a sharper, otherwise true, statement "proving" that the maximum of the left side is $1-\frac{1}{16}$. The essence of their reasoning was essentially the following:

"Since $\frac{1}{[x, y]} \leq \frac{1}{2 x}$, then $\frac{1}{[a, b]} \leq \frac{1}{2}$; the first term is therefore maximal when $a=1$ and $b=2$. Then $\frac{1}{[b, c]}=\frac{1}{[2, c]} \leq \frac{1}{4}$ and this is only possible if $c=4$. If $c=4$, then $\frac{1}{[c, d]} \leq \frac{1}{8}$ and thus $d=8$, and similarly $e=16$, so the maximum of the left side is $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=1-\frac{1}{16}$."

This reasoning is flawed, although the result is true. When we need to maximize something in several steps, we do not necessarily get the best result if we try to achieve the maximum possible increase at each step. In our case, it was tacitly assumed that the left side is maximal when even the first term is the largest possible. This determines the values of $a$ and $b$, and if we again "try to get as much as possible," this uniquely determines $c$, and so on. However, it is possible that if we are too greedy at the beginning, it will severely limit our options later.

A clear example illustrating the above is the following. Suppose the sum of two non-negative numbers is 1, and we want to maximize their product. The first factor is obviously the largest when it is 1, but then the second factor can only be 0, and the product is clearly not maximal.

Another example arises if we slightly modify the original problem. In the cited incorrect solutions, it was "proven" that under the conditions

$$\frac{1}{[a, b]}+\frac{1}{[b, c]}+\frac{1}{[c, d]}+\frac{1}{[d, e]}$$

is maximal when $a=1, b=2, c=4, d=8$ and $e=16$. If instead we seek the maximum of

$$\frac{1}{\sqrt{[a, b]}}+\frac{1}{\sqrt{[b, c]}}+\frac{1}{\sqrt{[c, d]}}+\frac{1}{\sqrt{[d, e]}}$$

then according to the cited "solution"

$$\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{8}}+\frac{1}{\sqrt{16}}=1.81066$$

is obtained, but this is not the maximum now, because if $a=1, b=2, c=3, d=6, e=12$, then the value of (3) is

$$\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{6}}+\frac{1}{\sqrt{6}}+\frac{1}{\sqrt{12}}=1.81228$$

It is clearly visible that the second term of the sum has decreased, but through this we were able to increase the third and fourth terms so that the sum exceeds the previous value.

2. We mentioned that a sharper statement is also true. We can see this by complete induction on $n$ ($n \geq 2$). Let $1 \leq a_{1}<a_{2}<\ldots<a_{n}$, $a_{i}$ be integers. Then we claim that

$$\frac{1}{\left[a_{1}, a_{2}\right]}+\frac{1}{\left[a_{2}, a_{3}\right]}+\ldots+\frac{1}{\left[a_{n-1}, a_{n}\right]} \leq 1-\frac{1}{2^{n}}$$

If $n=2$, then this is true by the estimate $\frac{1}{\left[a_{1}, a_{2}\right]} \leq \frac{1}{2 a_{1}} \leq \frac{1}{2}$.

Assume the statement is true for $n$, and let's show that it is also true for $(n+1)$. We distinguish two cases.

1. case: $a_{n+1}<2^{n+1}$. Then, by the estimate of the second solution,

$$\frac{1}{\left[a_{1}, a_{2}\right]}+\frac{1}{\left[a_{2}, a_{3}\right]}+\ldots+\frac{1}{\left[a_{n}, a_{n+1}\right]} \leq \frac{1}{a_{1}}-\frac{1}{a_{n+1}}<\frac{1}{a_{1}}-\frac{1}{2^{n+1}} \leq 1-\frac{1}{2^{n+1}}$$

2. case: $a_{n+1} \geq 2^{n+1}$. Clearly, $\left[a_{n}, a_{n+1}\right] \geq a_{n+1}$, so now $\frac{1}{\left[a_{n}, a_{n+1}\right]} \leq \frac{1}{2^{n+1}}$. Using the induction hypothesis,

$$\begin{aligned} \left(\frac{1}{\left[a_{1}, a_{2}\right]}\right. & \left.+\frac{1}{\left[a_{2}, a_{3}\right]}+\ldots+\frac{1}{\left[a_{n-1}, a_{n}\right]}\right)+\frac{1}{\left[a_{n}, a_{n+1}\right]} \leq \\ & \leq\left(1-\frac{1}{2^{n}}\right)+\frac{1}{2^{n+1}}=1-\frac{1}{2^{n+1}} \end{aligned}$$

This completes the proof. From the proof, it is also clear that in (4) equality holds precisely when $a_{i}=2^{i-1}$, $i=1,2, \ldots, n$.