Problems with unexpectedly short solutions
This was inspired by the following problem on quora:
Given that real non-zero numbers $a, b, c, d$ satisfy $^2+11^2+21^2+27^2 \le 6(+2+3) $, How would you prove that $a, b, c, d$ are consecutive terms of a geometric progression?
I hadn't seem this type of problem before, but after a while, I came up with this solution:
If $f(a, b, c, d) =^2+11^2+21^2+27^2 - 6(+2+3) $ then $ f(a, b, c, d)= (a-3b)^2+2(3c-b)^2+3(3d-c)^2 $. Therefore, if $f(a, b, c, d) \le 0$ then $a=3b, b=3c, c=3d $ so $a=27d, b=9d, c=3d$.
In case you want to generalize this, the $6$ on the right comes from $6 = 2\cdot 3$, and the $1, 2, 3$ on the right comes from the coefficients of the squared terms.
Anyway, the brevity of this solution struck me, and I got to wondering what other problems had unexpectedly short solutions.
So that's my question: what other problems have unexpectedly short solutions.
I'll start things off with this one:
Show that $n^{1/n} \to 1$ as integral $n \to \infty$.
Answer:
By Bernoulli's inequality, $(1+\frac1{\sqrt{n}})^n \ge 1+\frac{n}{\sqrt{n}} \gt \sqrt{n} $. Raising to the $2/n$ power, $n^{1/n} \lt (1+\frac1{\sqrt{n}})^2 \lt 1+\frac{3}{\sqrt{n}} \to 1 $.
Your turn.