Let $p$ and $q$ be positive numbers satisfying $\dfrac1p+\dfrac1q= 1$; and let $f ∶ [0,+∞) \to \mathbb{R}$ be the function $f(x) =\dfrac1p x^p− x +\dfrac1q$
Show that $f$ has an absolute minimum at $x = 1$ and hence deduce the inequality $ab \le \dfrac1p a^p + \dfrac1q b^q$ for any positive $a$ and $b$. Hint. What can you say on the magnitude of $p$ and $q$?
I can find the absolute minimum at $x=1$ but I can't establish the inequality. Please help!!
Since $f$ attains its minimum value at $x = 1$, and $f(1) = 0$, we look for a positive number to be substituted into $f$ so as to derive the required inequality, which is known as Young's inequality.
Consider $x = \dfrac{a}{b^\frac{q}{p}}$.
$f(x) \ge f(1)$
$\dfrac1p \left(\dfrac{a}{b^\frac{q}{p}}\right)^p - \left(\dfrac{a}{b^\frac{q}{p}}\right) + \dfrac1q \ge 0$
Note that $1-q = -\dfrac{q}{p} \iff p-pq = -q \iff p + q = pq \iff \dfrac1p+\dfrac1q=1$, so $b^{-\frac{q}{p}}=b^{1-q}$
$\dfrac1p a^p b^{-q} - ab^{1-q} + \dfrac1q \ge 0$
Multiply $b^{-q}$ to both sides, and we get the required inequality.