# 08 S2 Q3

Having seen the solution on the integral math site, my contradictions don't line up with theirs, but I assume that there are multiple ways it could be done.

For part (i) Forming inequalities for all numbers then expanding the brackets you obtain $ab-abc>\frac{4}{27}$ , $ac-abc>\frac{4}{27}$ and $bc-abc>\frac{4}{27}$

If you subtract any one of the inequalities from the other you obtain something of the form $a(b-c)>0$ which without loss of generality, can apply for any permutation of a, b and c.

In the case that a, b and c are distinct, then three inequalities will have to be less than zero as the bracket becomes negative ie if say $b\frac{1}{4}$ and $q(1-p)>\frac{1}{4}$
So you can obtain $p -q > 0$ and $q-p> 0$
From which it is even clearer to see that the inequalities cannot hold so by contradiction the statement holds.

This feels less elegant than the proof the answers have so would it have less marks given to it if something like this happened in the exam? - Unless I've actually made a mistake in my workings somewhere

### Somehow I messed up my latex

Somehow I messed up my latex so it's missing the middle part of the proof
In the case that a, b and c are distinct, then three inequalities will have to be less than zero as the bracket becomes negative,
if say $b< c$ then $a(b-c)<0$ since a is positive.
In the case that any of a, b or c are not distinct then two brackets will become zero so the inequality does not hold there either. Therefore by contradiction, one of the numbers must be less than or equal to $\frac{4}{27}$

A similar argument applies for part (ii)
$p(1-q)>\frac{1}{4}$ and $q(1-p)>\frac{1}{4}$
So you can obtain $p -q > 0$ and $q-p> 0$

### It seems alright, you wouldn

It seems alright, you wouldn't lost any marks since it's a proper proof by contradiction and fits all the requirements. That said, I'm sure you can see it's unnecessarily convoluted, especially as the former part hints very heavily at a certain method.

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