= Solution
Since
$$
f''(x)=(a-1)x^{a-2}>0
$$
for $x>0$, the second-derivative criterion shows that $f(x)=x^a/a$ is <strictly convex function>[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> is
$$
p=f'(x)=x^{a-1},
\qquad x=p^{1/(a-1)}.
$$
It is the unique global maximizer of $px-f(x)$, and hence
$$
\boxed{f^*(p)=\frac{p^b}{b}},
\qquad p>0.
$$
Applying the same calculation with the conjugate exponent $b$ gives
$$
\boxed{(f^*)^*(x)=\frac{x^a}{a}=f(x)},
\qquad x>0.
$$
Under the <convex conjugate>[Legendre-Fenchel transform] on all real dual variables, the same supremum additionally gives $f^*(p)=0$ for $p\leq0$.
Now take $a=r$, $b=s$, and apply the <Fenchel–Young inequality> $f(x)+f^*(y)\geq xy$. This yields <Young inequality>[Young's inequality for products]
$$
\boxed{\frac{x^r}{r}+\frac{y^s}{s}\geq xy}
$$
for positive $x,y$ and conjugate exponents $r,s$.
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