Showing when Young's Inequality is in fact equality.

Here is an alternative argument for the equality condition:

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Young's Inequality: $\forall p\in]1,\infty[, \frac{1}{p}+\frac{1}{q}=1,\forall a,b\in]0,\infty[: ab\leq\frac{a^p}{p}+\frac{b^q}{q}$. Furthermore

$$ab=\frac{a^p}{p}+\frac{b^q}{q} \iff a^p=b^q.$$

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Observation: \begin{align} a^p=b^q \iff & a^{p/q}=b \iff & a^{p-1}=b \\ & a=b^{q/p} & a=b^{q-1} \end{align}

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Lemma: Set

\begin{align} f:&]0,\infty[\to\Bbb{R} \\ &x\mapsto (1-x^{1-p})+(p-1)(1-x). \end{align} Then $f(x)=0 \iff x=1.$

Proof of Lemma: First observe that $f(1)=0$. Also $f'(x)=(p-1)(x^{-p}-1)$, from which we deduce that $1$ is the maximum point of $f$. $\checkmark$

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Proof of the Equality Condition:

$(\impliedby)$ First suppose $a^p=b^q$. Then by the observation above

\begin{align} \dfrac{a^p}{p}+\dfrac{b^q}{q}=\dfrac{(q+p)b^q}{pq}=b^q=b^{q-1}b=ab. \end{align}

$(\implies)$ Conversely suppose we have equality. Then

\begin{align} &ab=\dfrac{a^p}{p}+\dfrac{b^q}{q} =\dfrac{qa^p+pb^q}{p+q}\\ &\implies p+q =q\dfrac{a^{p-1}}{b} + p\dfrac{b^{q-1}}{a}\\ &\implies q\left(\dfrac{a^{p-1}}{b}-1\right)=p\left(1-\dfrac{b^{q-1}}{a}\right)\\ &\implies \left(\dfrac{a^{p-1}}{b}-1\right)=\dfrac{p}{q}\left(1-\dfrac{b^{q-1}}{a}\right)\\ &\implies \left(\left(\dfrac{b^{q-1}}{a}\right)^{1-p}-1\right)=(p-1)\left(1-\dfrac{b^{q-1}}{a}\right)\\ &\implies 0=\left(1-\left(\dfrac{b^{q-1}}{a}\right)^{1-p}\right)+(p-1)\left(1-\dfrac{b^{q-1}}{a}\right) \end{align}

Hence $\dfrac{b^{q-1}}{a}$ is a root of $f$ in the above lemma, which guarantees that $\dfrac{b^{q-1}}{a}=1$. By the observation this is equivalent to $a^p=b^q$.

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Note: I used certain arithmetical properties of the conjugates $(p,q)$. I'll leave those and the observation for you to check.

Your Exercise is wrong. Equality holds if and only if $a^p = b^q$.

To see this, consider $\def\e{\mathrm{e}}x \mapsto \e^x$, which is strictly convex on $\mathbb{R}$. Then, we get \begin{equation*} a \, b = \e^{\ln(a) + \ln(b)} = \e^{p^{-1} \, \ln(a^p) + q^{-1} \, \ln(b^q)} \le p^{-1}\, \e^{\ln(a^p)} + q^{-1} \, \e^{\ln(b^q)} = \frac{a^p}{p} + \frac{b^q}{q}. \end{equation*} From this, you get the characterization of equality.