On an inequality connecting polynomials and their derivatives I am currently reading Soundararajan's survey paper on GPY method. On page 14, the author claims that if $Q(x)$ is a non-constant polynomial such that $Q(0)=0$, then for integer $k\ge2$ the following inequality holds
$$
\int_0^1 x^{k-2}[Q(1-x)]^2\mathrm dx<{4\over k(k-1)}\int_0^1x^{k-1}[Q'(1-x)]^2\mathrm dx.
$$
However, the author did not give a proof of this result and instead asked the reader to find it out.
Here's my attempt: I first set $Q(x)=xF(x)$, where $F(x)$ can be any polynomial, so we have $Q'(x)=xF'(x)+F(x)$. Then, it follows from $(a+b)^2\ge2ab$ there is
$$
\int_0^1x^{k-1}[Q'(1-x)]^2\mathrm dx\ge2\int_0^1(1-x)^{k-1}xF(x)F'(x)\mathrm dx,
$$
but I don't know how to proceed further.
 A: In fact, the inequality is valid as long as $Q(x)$ is continuously differentiable. By definition, we have
$$
Q(1-t)=\int_t^1Q'(1-u)\mathrm du.
$$
Thus, it follows from Cauchy-Schwarz inequality that for all $\alpha>1$,
$$
[Q(1-t)]^2\le\int_t^1u^\alpha[Q'(1-u)]^2\mathrm du\cdot{t^{1-\alpha}-1\over\alpha-1}
$$
Plugging this inequality back into the left hand side, we get
\begin{aligned}
\int_0^1t^{k-2}[Q(1-t)]^2
&\le\int_0^1u^\alpha[Q'(1-u)]^2\int_0^ut^{k-2}{t^{1-\alpha}-1\over\alpha-1}\mathrm dt\mathrm du \\
&=\int_0^1u^{k-1}[Q'(1-u)]^2h(u)\mathrm du,
\end{aligned}
in which
$$
h(u)={u\over\alpha-1}\left[{1\over k-\alpha}-{u^{\alpha-1}\over k-1}\right].
$$
Taking the derivative, we see that the maximum of $h(u)$ is attained whenever
$$
u^{\alpha-1}={k-1\over\alpha(k-\alpha)}.
$$
Consequently, when $k>\alpha+1$ we have $u<1$ and
$$
h(u)<{1\over\alpha-1}\left(1-\frac1\alpha\right){1\over k-\alpha}={1\over\alpha(k-\alpha)},
$$
and when $k\le\alpha+1$, we see that the maximum of $h(x)$ in $[0,1]$ is attained at $x=1$:
$$
h(1)={1\over\alpha-1}\left({1\over k-\alpha}-{1\over k-1}\right)={1\over(k-\alpha)(k-1)}.
$$
Finally, setting $\alpha=(k+1)/2$ gives us
$$
\max_{0\le x\le1}h(x)\le{4\over k(k-1)}
$$
for all $k\ge2$. Finally, because $h(0)=0$, we see that there exists $\varepsilon>0$ such that $h(x)$ is strictly less than $4/k(k-1)$ in $[0,\varepsilon]$, so we obtain the final inequality:
\begin{aligned}
\int_0^1u^{k-1}[Q'(1-u)]^2h(u)\mathrm du
&\le{4\over k(k-1)}\int_\varepsilon^1u^{k-1}[Q'(1-u)]^2\mathrm du+\int_0^\varepsilon u^{k-1}[Q'(1-u)]^2h(u)\mathrm du \\
&<{4\over k(k-1)}\int_0^1u^{k-1}[Q'(1-u)]^2\mathrm du.
\end{aligned}
