Showing when Young's Inequality is in fact equality. 
The ellipses is where I'm stuck. I don't think a simple algebraic manipulation will work here.
 A: Here is an alternative argument for the equality condition:

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

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}

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 $f$. $\checkmark$

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 $a^p=b^q$.

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