Re: prove 7>(2^0.5+5^0.5+11^0.5)
@Bacterius:
In thinking about your logic going from bottom to top, I discovered that it's not the square roots that are the "problem". The rule for inequalities is this: any time you take a monotonically increasing function and apply it to both sides of an inequality, you get to preserve the direction of the inequality. Any time you take a monotonically decreasing function and apply it to both sides of an inequality, you need to reverse the direction of the inequality. If the function is not monotonically increasing or decreasing, then watch out!
The square root function is monotonically increasing on its domain. So, no problem. The square function, on the other hand, is monotonically decreasing for negative inputs, and monotonically increasing for positive outputs. So any time you square something in an inequality proof, you need to check that both sides are positive (keep the direction of the inequality) or that they are both negative (reverse the inequality).
Now, if I start at the bottom of your proof and work my way up, I notice that there are only three different kinds of steps in your proof: rewrites, square roots, and "shuffles" (shifting things from one side to the other by using addition or subtraction, which never requires an inequality reversal). You never square anything. So I think your proof is fine.
I think in my mind I was wondering about square roots, because if you start from $x^{2}>y^{2}$, you could either have $x>y$ or $-x<-y$. But in your proof, there are no variables anywhere, it's all numbers. And if you start from the bottom and work your way up, you can see that you take the positive square root a number of times, which is perfectly fine.