Let a and b be positive real numbers. If x^2 + y^2<=1 then what is the largest that ax+by can get?
By the Cauchy-Schwartz inequality, we have (x2 + y2)(a2+b2) >= (ax+by)2.
This can be transformed into (ax+by)2 <= (x2 + y2)(a2+b2) <= 1 * (a2+b2) <= (a2+b2). Hence ax + by <= sqrt(a^2 + b^2) and the equality is achieved when there exists ay = bx.
TD
Related MAT University answers
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