= Normalization proof of Young's inequality
{title2=$A=x^p/p+y^q/q,\quad X=x/A^{1/p},\quad Y=y/A^{1/q}$}
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 $XY\leq1$. Since $1/p+1/q=1$, this means $xy/A\leq1$, proving $xy\leq x^p/p+y^q/q$. 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.
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