How Does Proof by Contradiction Validate a/b + b/a >= 2?

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[SOLVED] basic math proof by contradiction

Homework Statement



prove: If a and b are positive numbers, a/b +b/a>=2

Homework Equations





The Attempt at a Solution



by contradiction (a^2+b^2)/ab<2 and got lost
 
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If (a^2+b^2)/ab<2 then a^2+b^2<2ab. So a^2+b^2-2ab<0. But a^2+b^2-2ab=(a-b)^2<0. What could be wrong with that?
 
thanks it was that easy
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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