I need puzzles

Post your math puzzles!

At what level?

Prove that 0 ≤ yz + zx + xy - 2xyz ≤ 7/27 where x, y and z are non-negative real numbers for which x + y + z = 1.

No No, something like..

You have 12 coins and a scale. 1 is fake and weighs differently than the rest. In no more than 3 weighs, determine which coin is the fake.

I can only think of easy ones, my middle school math teacher would always write these things on the board.

Theres that one from die hard, where you have a 3 gallon bucket and a 5 gallon bucket, and you have to fill a tub with exactly 4 gallons (or something like that).

Then theres the one where three guys check into a room, and through some weird mishap with their change and the bellboy's tip some of the money just seems to vanish in the math.  I don't remember the details of that one though.

A team of eight swimmers wants two enter two 4x100meter medley relays in a swim meet.  (The medley relay consists of one leg butterfly, one leg backstroke, one leg breaststroke, one leg freestyle.) Each swimmer can do all four strokes.

Keeping in mind that one swimmer can do only one leg, and that all four legs of both relays must be filled, devise an algorithm to allocate the swimmers so that the two relays have the lowest possible cumulative time.

>>4

You need additional info, such as: the fake coin is heavier or lighten than all others.

>>8 is right.  As of current, >>4 is impossible.

>>9 wrong

>>10 wrong

it's heavier, ok?

>>9 right

>>3
(1 - 2x)(1 - 2y)(1 - 2z) = 1 - 2(x + y + z) + 4(yz + zx + xy) - 8xyz = 4(yz + zx + xy) - 8xyz - 1. Hence yz + zx + xy - 2xyz = 1/4 (1 - 2x)(1 - 2y)(1 - 2z) + 1/4. By the arithmetic/geometric mean theorem (1 - 2x)(1 - 2y)(1 - 2z) ≤ ((1 - 2x + 1 - 2y + 1 - 2z)/3)3 = 1/27. So yz + zx + xy - 2xyz ≤ 1/4 28/27 = 7/27.

I r not smart.
http://www.kalva.demon.co.uk/imo/isoln/isoln841.html

weigh 4 on each side. if they are equal then the fake is in the last set of four. continue with sets of two then one. if one of the initial sets of four is off, choose that for sets of 2 then one last comparison.

as for the xyz problem. do a langrange multiplier maximization, where your first function is f(xy)=0 the constraint is g(xyz)=0.
set f'-(lambda)g'=0 then solve for lambda. notice that by this equation, once f is at a max/min, f' is zero, and so must g' be zero, meaning g is also at a max or min. the lambda will prolly be the latter fraction. you might be able to use matrices for this guy too, but that'd take more handwork.

swimming is gay

>>7
Do your own computer science homework.  Anyways without any differentiation between the swimmers, all times are rated equally, so it doesn't matter in which order the swimmers are chosen, and so choosing swimmer number 1 arbitrarily, and leg 1 arbitrarily:

N1 = {1, 2, 3, 4}
N2 = {5, 6, 7, 8}

Or in general

Set N (defining a set w/ 4 elements)
Nn = {An, An-1, An-2, An-3}
Where A is a set of all swimmers, with n defining the nth element of the set.

maybe.