prove that $[nx] \ge [x]+ \frac{[2x]}{2}+.....\frac{[nx]}{n} $
[.] is the floor function; $x\in R^+$
USAMO 1981.
So i posted this question earlier cause I couldn't find an 'understandable' solution anywhere, either they were wrong, or contained calculus or very bashy stuff.
I believe i have found a really nice solution using APMO 1999 / P2 (http://www.apmo-official.org/static/problems/apmo1999_prb.pdf).
So here goes,
NOTE THAT;
-[ia] + -[ja] $\ge$ -[ia+ja]
So we essentially need to prove
$-[nx] \le -[x]+ -\frac{[2x]}{2}+.....-\frac{[nx]}{n} $
Now using strong induction for all n $\le$ k, you know the drill, case for n=1,2 is trivial. assume it is true for all
n $\le k$
for some (k $\ge$ 2)
$-[x] \ge -[x]$
$-[x] + -\frac{[2x]}{2} \ge -[2x] $
.
.
.
$-[x]+ -\frac{[2x]}{2}+.....-\frac{[nx]}{n} \ge -[nx] $
adding all this stuff
-k[x] + -(k+1)$\frac{[2x]}{2}$ ....+ -$\frac{[nx]}{n}$ $\ge$ -[x] + -[2x] +..... -[kx]
Add -[x] + -[2x] +..... -[kx] to both sides and pairing [ia] and [ka-ia]on RHS
$(k+1)(-[x]+ -\frac{[2x]}{2}+.....-\frac{[nx]}{n})$ $\ge$ -[x]+ -[kx] + -[2x] + -[kx-x] +.... -[kx] + -[x]
$\ge -k[kx]$
After some very easy simplifying,
[(k+1)x] $\ge [x] + \frac{[2x]}{2} + ...... \frac{[(k+1)x]}{k+1} $
I'll be honest, i could do this only as I had seen that APMO question but i believe this is a nice method which i couldn't find anywhere on google.
If there is any mistake, do let me know