Normalization proof of Young's inequality ($A=x^p/p+y^q/q,X=x/A^{1/p},Y=y/A^{1/q}$)

The conjugate-power constrained product maximum proves Young's inequality for products by rescaling. For nonnegative $x,y$ and conjugate exponents $p,q>1$, let $A=x^p/p+y^q/q$. If $A=0$, both variables vanish. Otherwise the displayed normalized variables satisfy $X^p/p+Y^q/q=1$, hence $XY\le 1$. Since $1/p+1/q=1$, this means $xy/A\le 1$, proving $xy\le x^p/p+y^q/q$. Equality in the constrained maximum requires $X=Y=1$, equivalent to $x^p=y^q$; the zero case gives the same equality criterion. This approach simultaneously gives the sharp constant and every equality case.