So we were asked to solve a question in class about proof by contradiction...
Q) Suppose integers $1,2,3,\dots,10$ are placed randomly in a circular wheel. Show that the sum of any three consecutive integers is at least $15$.
Logical Answer: NO, because $1+2+3 = 6 < 15$.
Textbook Answer: PROVED. How?....
Let $A_r$ be the number positioned in the wheel at $r$-th position. Equation Set 1:
$\begin{align*} &A_1+A_2+A_3 \ge 15\\ &A_2+A_3+A_4 \ge 15\\ &A_3+A_4+A_5 \ge 15\\ &\qquad\qquad\qquad\vdots\\ &A_{10}+A_1+A_2 \ge 15 \end{align*}$
So, continuing proof by contradiction, lets assume Equation Set 1 is NOT true. Which implies--
Equation Set 2-
$\begin{align*} &A_1+A_2+A_3 < 15\\ &A_2+A_3+A_4 < 15\\ &A_3+A_4+A_5 < 15\\ &\qquad\qquad\qquad\vdots\\ &A_{10}+A_1+A_2 < 15 \end{align*}$
Now, adding all equations in Equation Set 2 we get
$3(A_1+\cdots+A_{10}) < 15\cdot10$
$\frac{3n(n+1)}2 < 150\;,$ where $n=10$ (cuz sum of $n$ integers is $n(n+1)/2$)
$3\cdot55 < 150$
$165 < 150$
Which is FALSE and is a contradiction to what we assumed (Equation set 2). Therefore our assumption is wrong and Equation Set 1 holds TRUE.
Which means sum of $3$ nos should be at least $15$. BUT logically thinking, a simple example of $1+2+3$ does not satisfy. What is the problem? Please Help!