Problem. Let $P(z)=\sum_{k=1}^{n}a_{k}z^{k}$ be a polynomial which is real on the real axis. Prove the inequality
$$\sum_{j=1}^{n}\sum_{k=1}^{n}\dfrac{a_{j}a_{k}}{1+j+k}=\int_{0}^{1}P(x)^{2}dx\leq\pi\sum_{j=1}^{n}\left|a_{j}\right|^{2}$$
by integrating $\log(z)P(z)^{2}$ over a suitable contour.
The equality is easy. My problem is getting the factor of $\pi$ on the right-hand side of the inequality; I keep getting $2\pi$. Here's what I have so far.
Let $\log z$ denote the branch of the logarithm with $0\leq \arg(z)<2\pi$. For $1\gg\epsilon,\delta>0$, consider the contour $\gamma$ given by
\begin{align*} &\gamma_{1}(t):=e^{2\pi it}, \qquad \epsilon\leq t\leq 1-\epsilon\\ &\gamma_{2}(t):=(1-t)e^{2\pi i(1-\epsilon)},\qquad 0\leq t\leq 1-\delta\\ &\gamma_{3}(t):=\delta e^{-2\pi i t},\qquad \epsilon\leq t\leq 1-\epsilon\\ &\gamma_{4}(t):=te^{2\pi i\epsilon},\qquad \delta\leq t\leq 1 \end{align*}
Since $f(z):=\log(z)P(z)^{2}$ is holomorphic on and insider the contour, we see from Cauchy's theorem that
$$0=\int_{\gamma}\log(z)P(z)^{2}dz=\left(\int_{\gamma_{1}}+\int_{\gamma_{2}}+\int_{\gamma_{3}}+\int_{\gamma_{4}}\right)f(z)dz$$
I get that \begin{align*} \int_{\gamma_{1}}f(z)dz&=\int_{\epsilon}^{1-\epsilon}(2\pi i t)P(e^{2\pi i t})^{2}(2\pi i)e^{2\pi i t}dt\\ &\rightarrow-4\pi^{2}\int_{0}^{1}P(e^{2\pi i t})^{2}e^{2\pi i t}dt, \ \epsilon\rightarrow 0^{+} \end{align*}
\begin{align*} \int_{\gamma_{3}}f(z)dz&=\int_{\epsilon}^{1-\epsilon}\left[\log\delta+2\pi i t\right]P(\delta e^{2\pi i t})^{2}(2\pi i)\delta e^{2\pi i t}dt\\ &\rightarrow 0, \ \delta\rightarrow 0^{+} \end{align*} by dominated convergence. Now observe that
\begin{align*} \int_{\gamma_{4}}f(z)dz&=\int_{\delta}^{1}\left[\log\left|t\right|+2\pi i\epsilon\right]P(te^{2\pi i\epsilon})^{2}e^{2\pi i\epsilon}dt\\ &\rightarrow\int_{\delta}^{1}\log\left|t\right|P(t)^{2}dt, \ \epsilon\rightarrow 0^{+}\\ &\rightarrow\int_{0}^{1}\log\left|t\right|P(t)^{2}dt, \ \delta\rightarrow 0^{+} \end{align*} by dominated convergence. Similarly,
\begin{align*} \int_{\gamma_{2}}f(z)dz&=\int_{0}^{1-\delta}\left[\log\left|1-t\right|+2\pi i(1-\epsilon)\right]P((1-t)e^{2\pi i(1-\epsilon)})^{2}-e^{2\pi i(1-\epsilon)}dt\\ &\rightarrow-\int_{0}^{1-\delta}\left[\log\left|1-t\right|+2\pi i\right]P(1-t)^{2}dt, \ \epsilon\rightarrow 0^{+}\\ &\rightarrow-\int_{0}^{1}\left[\log\left|1-t\right|+2\pi i\right]P(1-t)^{2}, \ \delta\rightarrow 0^{+} \end{align*} again by dominated convergence. Making the change of variable $s=1-t$, we obtain that \begin{align*} \lim_{\delta\rightarrow 0^{+}}\lim_{\epsilon\rightarrow 0^{+}}\int_{\gamma_{2}}f(z)dz&=\int_{1}^{0}\left[\log\left|s\right|+2\pi i\right]P(s)^{2}ds=-\int_{0}^{1}\left[\log\left|s\right|+2\pi i\right]P(s)^{2}ds \end{align*}
Combining these results, we see that $$0=\lim_{\delta\rightarrow 0^{+}}\lim_{\epsilon\rightarrow 0^{+}}\int_{\gamma}f(z)dz=-4\pi^{2}\int_{0}^{1}P(e^{2\pi i t})^{2}e^{2\pi it}dt-2\pi i\int_{0}^{1}P(t)^{2}dt$$
Since $P$ is real on the real axis $P(t)^{2}=\left|P(t)\right|^{2}$, for $0\leq t\leq 1$, whence \begin{align*} 2\pi\int_{0}^{1}P(t)^{2}dt\leq4\pi^{2}\int_{0}^{1}\left|P(e^{2\pi i t})^{2}\right|dt&=\int_{0}^{1}P(e^{2\pi i t})\overline{P(e^{2\pi i t})}dt\\ &=4\pi^{2}\sum_{j,k=1}^{n}a_{j}\overline{a_{k}}\int_{0}^{1}e^{2\pi(k-j)it}dt\\ &=4\pi^{2}\sum_{j=1}^{n}\left|a_{j}\right|^{2}, \end{align*} by the orthogonality relations.
In your limit for the integral over $\gamma_1$, you dropped a $t$ coming from the logarithm. With that included, after taking limits, you have
\begin{align} \int_0^1 P(t)^2\,dt &= 2\pi i \int_0^1 tP(e^{2\pi it})^2 e^{2\pi it}\,dt\\ &\leqslant 2\pi \int_0^1 t\lvert P(e^{2\pi it})\rvert^2\,dt\\ &= 2\pi\sum_{j,k=1}^n a_ja_k\int_0^1 t e^{2\pi i (j-k)t}\,dt\\ &= \pi\sum_{j=1}^n \lvert a_j\rvert^2 + 2\pi\sum_{j\neq k} a_j a_k\int_0^1 t e^{2\pi i(j-k)t}\,dt\\ &= \pi \sum_{j=1}^n \lvert a_j\rvert^2 + 4\pi \sum_{j > k} a_j a_k \int_0^1 t\cos (2\pi(j-k)t)\,dt.\tag{1} \end{align}
Now, for $m > 0$, we have
\begin{align} \int_0^1 t\cos (2\pi mt)\,dt &= \biggl[ \frac{t}{2\pi m}\sin (2\pi mt)\biggr]_0^1 - \frac{1}{2\pi m}\int_0^1 \sin (2\pi mt)\,dt\\ &= 0, \end{align}
which gives the desired inequality when inserted into $(1)$.