Inequality between integral of $(1-x)^k P(x)^2$ and integral of $(1-x)^{k+1} P'(x)^2$ Given a polynomial $P$ with real coefficients and an integer $k \ge 0$, define
$$I_k(P) = \int_0^1 \frac{(1-x)^k}{k!} P(x)^2 dx.$$
I would like to show that if $P(0) = 0$ and $P$ is not identically zero, then
$$I_k(P) \le \frac{4}{k+2} I_{k+1}(P').$$

The result is true for monomials $P(x) = x^r$. Indeed, we have
$$I_k(P) = \frac{(2r)!}{(k+2r+1)!}
\quad \text{and} \quad
I_{k+1}(P') = \frac{r^2(2r-2)!}{(k+2r)!},$$
so
$$\frac{I_{k-2}(P)}{I_{k-1}(P')} = \frac{2(2r-1)}{r(k+2r+1)} < \frac{4}{k+2}.$$

In terms of partial progress, I can show $I_k(P) \le 2 \sqrt{I_{k+1}(P) I_{k+1}(P')}$ and $I_k(P) \le \frac{1}{4k} I_k(P')$. Together, these give $I_k(P) \le \frac{1}{\sqrt{k+1}} I_{k+1}(P')$ (e.g. proving the result for $k \le 12$).
For the first, we may integrate by parts and then apply Cauchy-Schwarz to find
$$I_k(P) = 2 \int_0^1 \frac{(1-x)^{k+1}}{(k+1)!} P(x) P'(x) dx
\le 2 \left(\int_0^1 \frac{(1-x)^{k+1}}{(k+1)!} P(x)^2 dx\right)^{1/2}
\left(\int_0^1 \frac{(1-y)^{k+1}}{(k+1)!} P'(x)^2 dx\right)^{1/2}
= 2 \sqrt{I_{k+1}(P) I_{k+1}(P')}.$$
For the second, we may replace $P(x)$ with $\int_0^x P'(t) dt$ and then apply Cauchy-Schwarz to find
\begin{aligned}
I_k(P) &= \int_{0 \le t \le x \le 1} \frac{(1-x)^k}{k!} P(x) P'(t) \; dx \; dt \\
&\le \left(\int_{0 \le t \le x \le 1} \frac{(1-x)^{k+1}}{k!} P(x)^2 \; dx \; dt\right)^{1/2}
\left(\int_{0 \le t \le x \le 1} \frac{(1-x)^{k-1}}{k!} P'(t)^2 \; dx \; dt\right)^{1/2} \\
&= \left(\int_0^1 x(1-x) \frac{(1-x)^k}{k!} P(x)^2 \; dx\right)^{1/2}
\left(\int_0^1 \frac{(1-t)^k}{k \cdot k!} P'(t)^2 \; dt\right)^{1/2} \\
&\le \left(\frac{1}{4} I_k(P)\right)^{1/2} \left(\frac{1}{k} I_k(P')\right)^{1/2},
\end{aligned}
using in the last step that $x(1-x) \le \frac{1}{4}$.

Using the inequality $I_k(P) \le 2 \sqrt{I_{k+1}(P) I_{k+1}(P')}$, it suffices to show $I_{k+1}(P) \le \left(\frac{2}{k+2}\right)^2 I_{k+1}(P')$.
Integrating by parts and applying Cauchy-Schwarz, we have
\begin{align}
I_{k+1}(P)
&= 2 \int_0^1 \frac{(1-x)^{k+2}}{(k+2)!} P(x) P'(x) dx \\
&\le 2 \left(\frac{k+3}{k+2}\right)^{1/2}
\left( \int_0^1 \frac{(1-x)^{k+3}}{(k+3)!} P(x)^2 dx \right)^{1/2}
\left( \int_0^1 \frac{(1-x)^{k+1}}{(k+1)!} P'(x)^2 dx \right)^{1/2} \\
&= 2 \left(\frac{k+3}{k+2}\right)^{1/2} I_{k+3}(P)^{1/2} I_{k+1}(P')^{1/2}.
\end{align}
Since $(1-x)^{k+3} \le (1-x)^{k+1}$ for $0 \le x \le 1$, we have $I_{k+3}(P) \le \frac{1}{(k+3)(k+2)}I_{k+1}(P)$, so altogether we have shown
\begin{align}
I_{k+1}(P)
&\le 2 \left(\frac{k+3}{k+2}\right)^{1/2} \left(\frac{1}{(k+3)(k+2)}I_{k+1}(P)\right)^{1/2} I_{k+1}(P')^{1/2} \\
&= \left(\frac{2}{k+2}\right) I_{k+1}(P)^{1/2} I_{k+1}(P')^{1/2}.
\end{align}
Squaring and dividing by $I_{k+1}(P)$ gives the desired $I_{k+1}(P) \le \left(\frac{2}{k+2}\right)^2 I_{k+1}(P')$.
