The premise is that any real number is non-negative when squared, so integration of a squared real function must yield a non-negative number. What is done next is that we now f(x)^2 is greater than a two-variable expression in a and b, so we try to find a maximum. It does not have to do with any particular values because of convexity or concavity of the two variable expression (e.g. if the minimum of a convex function is postivie, then the whole function is positive).
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Yes. The maximum of the RHS is not the minimum of the LHS.
Correct. The minimum of the LHS is at least the maximum of the RHS. This is more than you need.
What is the proven in the article is that the minimum of the LHS is at most 4 and at least 0.