$P=\sum_{k=0}^{n}c_{k}z^{k}$ if $\forall t\in[k].c_t\in\mathbb{R}\rightarrow\int_{-1}^{1}P^2\leq\pi\sum_{k=0}^{n}c_{k}^{2}$

Question

$P=\sum_{k=0}^{n}c_{k}z^{k}$ if $\forall t\in[k].c_t\in\mathbb{R}\rightarrow\int_{-1}^{1}P^2\leq\pi\sum_{k=0}^{n}c_{k}^{2}$ My try: it's clear that $\frac{1}{2\pi}\int_{0}^{2\pi}|P(e^{i\theta})|^{2}d\theta=\sum_{k=0}^{n}c_{k}^{2}$ so: $\int_{-1}^{1}P^{2}\leq2\pi\sum_{k=0}^{n}c_{k}^{2}$ I can't understand why the tighter bound exists