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> The longest inequality on that page is true for all a, b If you agree with this, then you why do also believe some particular a,b proves that ∫f²dx ≥ 4?

• mbustamanter 4y ago
  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).
  • yobbo 4y ago
    Yes. The maximum of the RHS is not the minimum of the LHS.
    • Smaug123 4y ago
      Correct. The minimum of the LHS is at least the maximum of the RHS. This is more than you need.
      • yobbo 4y ago
        What is the proven in the article is that the minimum of the LHS is at most 4 and at least 0.