Is this a Morera´s Theorem Application? Let $G \subset \mathbb C$ be a domain and $f: G \to \mathbb C$ a continuous function such that for any closed and rectifiable path $\gamma \subset G$, 
$$
\left| \oint_\gamma f(z)dz \right|\leq \left( L(\gamma)\right)^3.
$$
Where $L(\gamma)$ denote the arc length of the path $\gamma$, that is $L(\gamma)=\int_\gamma |dz|$. Prove that $f$ is holomorphic in $G$.
My Ideas: This looks like a good candidate to use the powerful result of Morera´s Theorem, which tells us that is enough to prove that $\oint_\gamma f(z)dz$ vanishes for any closed path $\gamma \subset G$. However I don't seem to put together a good argument and I am starting to believe this may be prove in another way. 
Why 3? Why use the $3$ in "$\leq \left( L(\gamma)\right)^3$"? I guess I would know that if I could prove it but do the assertion still holds if we replace $3$ by $2$, $5$, $8$ or any other whole number?
Any help on how proving this or hints will be very appreciated.
 A: Take any curve $\gamma$ that traces a rectangle.  Split the rectangle into $n^2$ smaller rectangles, traced by curves $(\kappa_i)_{1\le i \le n^2}$ whose lengths are $1/n$ times the length of $\gamma$.  Then
$$ \oint_\gamma f(z) \, dz = \sum_{i=1}^{n^2} \oint_{\kappa_i}  f(z) \, dz.$$
Go from there.
From this you see that the 3 can be replaced by any number strictly bigger than 2.  But 2 won't work.  This follows by seeing that for almost any small enough $f$ that is smooth, and by using the Gauss-Green formula, that the inequality holds with 2, because the area enclosed by a curve is always bounded its length squared times $1/4\pi$. 
A: Same idea of the other answer but this one has the flavor of Goursat's theorem proof. 
By Morera´s Theorem it suffices to show that for any triangular path $T \subset G$ ($T=\partial \Delta$ where $\Delta$ is a triangle), $\int_T f(z)dz=0$. Take any triangular path $T=\partial \Delta \subset G$. Now using the mid points of $\Delta$ form 4 triangles $\Delta_j$, $j=1,2,3,4$, such that each $T_j = \partial \Delta_j$ is a triangular path, take orientation on each boundary as shown in the next figure:

Clearly $L(T_j)=(1/2)L(T)$ and $\int_T=\sum_{j=1}^4\int_{T_j}$. Let $T^{(1)}$ be the path in  $\{T_1, \cdots T_4\}$ such that
$$
\left| \int_{T_j} f(z) dz \right| \leq \left|\int_{T^{(1)}} f(z) dz \right| \ \forall \ j=1,2,3,4
$$
Hence 
$$
\left| \int_{T} f(z) dz \right| \leq 4 \left|\int_{T^{(1)}} f(z) dz \right| 
$$
Doing the same thing, but now to  $\Delta^{(1)}= \partial T^{(1)}$ and so on, we obtain for each $n \in \mathbb{N}$ a triangular path $T^{(n)}$ such that $L(T^{(n)})=(1/2^n)L(T)$ and 
$$
\left|\int_{T} f(z) dz  \right| \leq 4^n \left|\int_{T^{(n)}} f(z) dz  \right|  \leq 4^n \left( L\left(T^{(n)}\right)\right)^3 = \frac{4^n}{(2^n)^3} \left( L\left(T\right)\right)^3
$$
However, since
$$
\lim_{n \to \infty} \frac{4^n}{(2^n)^3} = \lim_{n \to \infty} \frac{1}{2^n}=0 \tag{1}
$$
indeed $\int_{T} f(z) dz  =0$ for any triangular path $T \subset G$.
Note As it was told by @StephenMontgomery-Smith answers, this holds if you use any real number $t>2$, since the limit (1) stills goes to $0$ replacing $3$ for $t>2$.
