 Solution

Sincefor $x>0$, the second-derivative criterion shows that $f(x)=x^a/a$ is strictly convex. Put $b=a/(a-1)$, so $1/a+1/b=1$. For dual variable $p>0$, the stationary equation in the Legendre transform isIt is the unique global maximizer of $px-f(x)$, and henceApplying the same calculation with the conjugate exponent $b$ givesUnder the Legendre-Fenchel transform on all real dual variables, the same supremum additionally gives $f^{\ast }(p)=0$ for $p\le 0$.

Now take $a=r$, $b=s$, and apply the Fenchel–Young inequality $f(x)+f^{\ast }(y)\ge xy$. This yields Young's inequality for productsfor positive $x,y$ and conjugate exponents $r,s$.