Past exam of the short options course of the University of Oxford / 2015 / 4 / d / Solution

Use a normalization proof of Young's inequality. For arbitrary $x,y\ge 0$, putIf $A=0$, both coordinates are zero and the desired inequality is equality. If $A>0$, defineThen $X^p/p+Y^q/q=1$. Part (c) gives $XY\le 1$, and the conjugate exponent identity gives $XY=xy/A$. Therefore Young's inequality for products isThe unique equality case in part (c) says $X=Y=1$. Translated back, and including the zero case, this gives the sharp conditionConversely, if both powers equal $r\ge 0$, then $xy=r^{1/p+1/q}=r$ and the right side is $r(1/p+1/q)=r$, directly verifying equality.