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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.

The feasible set is closed and contained in the rectangleIt is therefore compact, and the continuous function $xy$ attains a global maximum there by the extreme value theorem. At the two boundary endpoints where a coordinate is zero, the product is zero. Meanwhile $(1,1)$ is feasible because the conjugate exponents satisfy $1/p+1/q=1$, and its product is one. Hence every maximizer has $x,y>0$.

On that positive region the constraint is $C^1$ with nonzero gradient. The Lagrange multiplier equations areThe first two identities imply $xy=\lambda x^p=\lambda y^q$. Since $x,y>0$, the multiplier is nonzero, so $x^p=y^q=r$. The constraint now givesThus the sole interior candidate is $(1,1)$. Existence of a maximum and the boundary comparison prove the unique global resultThis conjugate-power constrained product maximum is valid for all real $p,q>1$ with $1/p+1/q=1$; no integrality of the exponents is needed.