If 0 < n < m < 1, m^2 n^2 must be less than which of the following e

Take it as given that 0 < n < m < 1.

Let's start by factoring m^2 - n^2.
--> m^2 - n^2 = (m + n) * (m - n)

Can we say anything about the sign of this number? We know it's greater than zero. (Unfortunately this doesn't help, but is a good starting place for a question like this IMO)
--> We know that (m + n) > 0 because m > 0, n > 0.
--> We know that (m - n) > 0 because m > n
--> so (m + n) * (m - n) > 0

So let's proceed one by one until we find an expression we can prove.
--> A is easy to invent a counterexample for. I can pick m, n > 0.5 such that (m+n) > 1 ==> (m+n) * (m-n) > (m - n)
--> B doesn't look very promising ... instinctively, I want the proof to use my factored form right away ... skip and come back
--> C doesn't look very promising ... instinctively, I want the proof to use my factored form right away ... skip and come back
--> D looks promising. I'll try this one.
--> Didn't look at E.

Ok let's see if we can't prove D to be true.

We want to show m^2 - n^2 < (m + n)^2; I'll do this one by assuming the opposite and then forcing a contradiction.
--> Suppose it's true that (m+n)^2 <= m^2 - n^2
--> Start by expanding (m+n)^2: (m+n)^2 = (m+n) * (m+n)
--> We can then factor m^2 -n^2 into a similar form: m^2 - n^2 = (m+n) * (m-n)
--> Then we can rewrite our first inequality: (m+n)(m+n) <= (m+n)(m-n)
--> Because m, n > 0, we can just divide both sides by (m+n) safely: (m+n) <= (m-n)
--> Now take out m from both sides: n <= -n
--> And add n to both sides: 2n <= 0
--> And divide by 2: n <= 0
--> Note at the outset that we said n > 0. We've arrived at a contradiction.
--> So It cannot be true that (m+n)^2 <= m^2 - n^2
--> And therefore it must be true that (m+n)^2 > m^2 - n^2

D is the answer.