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.
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}