$H$ is a maximal normal subgroup of $G$ if and only if $G/H$ is simple. I need to prove that $H$ is a maximal normal subgroup of $G$ if and only if $G/H$ is simple.
My proof to the $(\Rightarrow)$ direction seems too much trivial:
Let us assume there exist $A$ so that $A/H\lhd G/H$. Then by definion, $H$ must be normal in $A$. Because $H$ is maximal, we get $H=A$ and therefore $A/H={1}$
Is it correct?
Update:
Now I see that I need to prove that not only $A\lhd H$ but also $A\lhd G$. Assumig I have proven that, is the proof correct?
 A: I flesh out the exquisite answer of B. S.
$\color{darkred}{ \text{ (1.)  If I'm not confounded, I think $H ⊴ G$ means $H ⊲ G$ or $H = G$. Is this perfect?  } }$
Forward step: $H \text{ maximal } ⊲ G \implies G/H$ simple.
Let $\frac{A}{H} ⊲\frac{G}{H}$  wherein $H ⊴A⊴G$. Since H is a maximal subgroup, $\begin{cases} H = A \implies \frac{A}{H}=1 \\ \text{ or } A=G \implies \frac{A}{H}=\frac{G}{H} \end{cases}$.
This means that $\frac{G}{H} $ is a simple group. ♥
Backward step: Now suppose that $H ⊲G$ and $\frac{G}{H} $ is simple.
If we have $H ⊴A⊴G$ then obviously $\frac{A}{H} ⊲\frac{G}{H}$.
By reason of the presupposition for this backward step, $\frac{G}{H}$ is simple.
Hence $\frac{A}{H}=\frac{G}{H}$  or $\frac{A}{H} =\{H\}$ . So, $A=G$ or $H=A$. ♥          
A: As you noted let $\frac{A}{H} \trianglelefteq \frac{G}{H}$  wherein $H \trianglelefteq A\trianglelefteq G.$ Since $H$ is a maximal subgroup, $H=A$ or $A=G$ and so, $\frac{A}{H}=1$ or $\frac{A}{H}=\frac{G}{H} $.
This means that $\frac{G}{H} $ is a simple group.
Now suppose that $H \trianglelefteq G$ and $\frac{G}{H} $ is simple. If we have $H \trianglelefteq A\trianglelefteq G$ then obviously $\frac{A}{H} \trianglelefteq\frac{G}{H}$  and that $\frac{G}{H}$ is simple, we get $\frac{A}{H}=\frac{G}{H}$  or $\frac{A}{H} =\{H\}$ . So, $A=G$ or $H=A$.
