Let $C = \{f: \mathbb{D} \to \mathbb{C} \ |\ f\text{ univalent},\ f(\mathbb{D})\text{ convex},\ f(0) = 0\text{ and } f'(0) = 1 \} $
I want to show that if $f \in C$ with the series expansion $f = z + a_2z^2 + a_3z^3+...$, then $|a_2^2-a_3| \leq \frac{1-|a_2|^2}{3}$
What I know is that if $f$ has the stated expansion, but not necessarily in $C$, then after considering $(f(z^{-1}))^{-1} = z - a_2 +(a_2^2-a_3)z^{-1} + ...$ with the area theorem, then $|a_2^2-a_3| \leq 1$. I can't think of anything else that relates to this. Please give me some clue.
Let $f\in C$. Then $ F(z)=zf^{\prime\prime}(z)/f^\prime(z) $ satisfies the conditions $$F(0)=0,\,\operatorname{Re} F(z)>-1 \,\,(|z|<1).$$ The linear transformation $$ w=\frac{z}{z+2} $$ maps $\{z|\operatorname{Re}z>-1\}$ onto $|w|<1$. So $g(z)=\frac{F(z)}{F(z)+2}$ satisfies $ g(0)=0$ and $ |g(z)|<1 \,\text{for } |z|<1.$ Therefore, by the lemma of Schwarz, $h(z)=g(z)/z$ is analytic and $|h(z)|<1$ in $|z|<1$.
Lemma: Let $h(z)$ be analytic in $|z|<1$ with $|h(z)|<1$. Then$$ |h^\prime(z)|\le \frac{1-|h(z)|^2}{1-|z|^2}$$ holds for $|z|<1.$ Especially we have \begin{align} |h^\prime(0)|\le 1-|h(0)|^2.\tag{1} \end{align} This is an easy corollary of the lemma of Schwarz.
If $f$ has the series expansion $$ f = z + a_2z^2 + a_3z^3+...\,, $$ then it is easy to show that $h(z)=\frac{f^{\prime\prime}}{zf^{\prime\prime}+2f^\prime}$ has the series expansion \begin{align} h(z)=a_2 + 3(a_3-a_2^2)z \, + ... \,. \end{align} By ($1$) we have$$ 3|a_3-a_2^2|\le 1-|a_2|^2,$$ which is the desired conclusion.