= Solution
Use a <normalization proof of Young's inequality>. For arbitrary $x,y\geq0$, put
$$
A=\frac{x^p}{p}+\frac{y^q}{q}.
$$
If $A=0$, both coordinates are zero and the desired inequality is equality. If $A>0$, define
$$
X=\frac{x}{A^{1/p}},\qquad Y=\frac{y}{A^{1/q}}.
$$
Then $X^p/p+Y^q/q=1$. Part (c) gives $XY\leq1$, and the <conjugate exponent> identity gives $XY=xy/A$. Therefore <Young's inequality for products> is
$$
\boxed{xy\leq\frac{x^p}{p}+\frac{y^q}{q}.}
$$
The unique equality case in part (c) says $X=Y=1$. Translated back, and including the zero case, this gives the sharp condition
$$
\boxed{\text{equality holds exactly when }x^p=y^q.}
$$
Conversely, if both powers equal $r\geq0$, then $xy=r^{1/p+1/q}=r$ and the right side is $r(1/p+1/q)=r$, directly verifying equality.
Back to article page
