Tungkol sa pagkakapantay-pantay $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$ // About this beautiful identity

 (Filipineza)

          The following Problem was published on Facebook_page Canadian Mathematical Society / Société mathématique du Canada : (in bilingual presentation, as usual)

A beautiful (but somewhat unfinished) solution was presented by Regragui El Khammal (الركراكي الخمال).

          My solution is the following.

ANSWER CiP

For all positive integers satisfying inequality

$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\;<\; 1 \tag{I}$$

the inequality in the statement 

$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\;\leqslant\;\frac{41}{42} \tag{ S}$$

 actually occurs. The sign "$=$" is obtained only for

 values $2,\; 3,\; 7\;$ for $a,\; b,\; c\;$ (in any order).

SOLUTION CiP

                       Let us first look for the triplets $(a,b,c)\in \mathbb{Z_+}\times \mathbb{Z_+}\times \mathbb{Z_+}$

 satisfying the inequality (I) and additionally verifying the condition

$$1\leqslant a \leqslant b \leqslant c \tag{C}\;.$$

Then the other triplets satisfying the inequality (I) are obtained by doing all permutations of those found.

               Let us then note the equality

$$\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1\;. \tag{E}$$               

          We will study three cases : I-III.

          I. If $a\geqslant 2$ and $b\geqslant 3$ then inequality (I) is satisfied for "sufficiently large" $c$.(Anyway $c \geqslant 7$, taking into account the equality (E); precisely $c \geqslant \left [\frac{1}{1-\frac{1}{a}-\frac{1}{b}}\right]+1$,$[.]$ being the floor function , but this precision is not necessary in solving.) In this case we have

$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leqslant \frac{1}{2}+\frac{1}{3}+\frac{1}{7}=\frac{41}{42}.$$

We have concluded (S). Equality is reached only for $a=2,\;b=3,\;c=7\;$, otherwise the above inequality is strict.

          II. The values $a=2\;$ and $b=2\;$ contradict inequality (I), so this case is not possible.

          III. Neither can $a=1\;$ because of the same inequality (I).

      Example: for $a=b=3$ we have $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{3}+\frac{1}{3}+\frac{1}{c}$ and this last expression is $<1$ for $c\geqslant 4$. But then

$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leqslant \frac{2}{3}+\frac{1}{4}=\frac{11}{12}=\frac{77}{84}=\frac{82}{82}=\frac{41}{42}\;;$$

we get a strict inequality.

$\blacksquare$