This is the last question! Thx!
Let a and b be positive real numbers such that a^2 + b^2 = 1. Prove that a^4 + b^4 >= 0.5
Edit: The proof below is invalid, as
But I'm leaving it here in case it sparks any other ideas.
Without loss of generality, we can let a ≥ b.
Note that if a² = .5, then b² = .5. Also note that if b² > .5, it must be that a² < .5. But this means that a < b, contradicting a ≥ b. Thus, we can let a² ≥ .5 ≥ b².
Since b^4 must be non-negative, a^4 + b^4 ≥ .5. QED
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
With the same reasoning from the previous proof, let:
a² = .5 + ∈, where ∈ ≥ 0.
Then a² + b² = .5 + ∈ + b² = 1. So b² = .5 - ∈. So:
Since ∈² ≥ 0, a^4 + b^4 ≥ .5. QED
Edit: Interesingly enough, given a^2 (or just a), you can easily calculuate exactly how much greater than .5 a^4 + b^4 will be, without ever having to know what b is or rasing anything to a 4th power.