If $f$ is T-periodic , continuous and piecewise $C^{1}([0,T])$ with $\displaystyle\int_{0}^{T}f(x)dx=0$
then we have $$\int_{0}^{T}|f(x)|^{2}dx\le\frac{T^{2}}{4\pi^{2}}\int_{0}^{T}|f'(x)|^{2}dx$$
My attempt :(edited)
Firstly , we define the $n^{th}$ Fourier coefficient of $f$ on the $[0,T]$ by $\widehat{f}(n)=\displaystyle\frac{1}{T}\int_{0}^{T}f(x)e^{-\frac{2\pi i nx}{T}}~dx$ and denote that $f\in L^{p}([0,T]),0<p<\infty~,$ if $\|f\|_{L^{p}([0,T])}=\bigg(\displaystyle\frac{1}{T}\int_{0}^{T}|f(x)|^{p}~dx\bigg)^{1/p}<\infty$ .
Now , as $f$ is piecewise $C^{1}$ on $[0,T]$ then $f'$ is Lebesgue integrable on $[0,T]$ and hence for all $n\in{\bf Z}-\{0\}$ one has , $$\widehat{f'}(n)=\frac{1}{T}\int_{0}^{T}f'(x)e^{-\frac{2\pi inx}{T}}dx=\frac{2\pi in}{T^{2}}\int_{0}^{T}f(x)e^{-\frac{2\pi inx}{T}}dx=\frac{2\pi in}{T}\widehat{f}(n)$$ , where the second equality used by the integration by parts and $f(0)=f(T)$ as $f$ is T-periodic .
Moreover , we have that $\widehat{f'}(0)=\displaystyle\frac{1}{T}\int_{0}^{T}f'(x)~dx=\frac{1}{T}\big(f(T)-f(0)\big)=0\color{blue}{~~-(1)}$ .
As $f$ is continuous on $[0,T]$ , one must have $f$ is Lebesgue integrable and bounded on $[0,T]$ so we must have $f\in L^{2}([0,T])$ since $L^{1}([0,T])\cap L^{\infty}([0,T])$ is embedded into $L^{2}([0,T])$ .
Hence , by Parseval's identity that \begin{align} \int_{0}^{T}|f(x)|^{2}dx&=T\sum_{n\in {\bf Z}}|\widehat{f}(n)|^{2}\\ &=T\sum_{n\ne 0}|\widehat{f}(n)|^{2}\\ &=\frac{T^{2}}{4\pi^{2}}T\sum_{n\ne 0}\frac{|{\widehat{f'}(n)}|^{2}}{\color{red}{n^{2}}}\\ &\le\frac{T^{2}}{4\pi^{2}}T\sum_{n\ne 0}|\widehat{f'}(n)|^{2}\\ &=\frac{T^{2}}{4\pi^{2}}T\sum_{n\in {\bf Z}}|\widehat{f'}(n)|^{2}\\ &=\frac{T^{2}}{4\pi^{2}}\int_{0}^{T}|f'(x)|^{2}dx \end{align} , where the second equality is according to the hypothesis that ${\widehat f}(0)=\displaystyle\frac{1}{T}\int_{0}^{T}f(x)dx=0$ and the fourth equality holds by the equation $\color{blue}{(1)}$ .
Can anyone check my working for validity if you the time , otherwise just skip the post . Any comment or valuable suggestion I will be grateful .
The red one is that I edited which is the suggestion by the user284331
A minor mistake is that, one should have $\displaystyle\sum_{n\ne 0}|\widehat{f}(n)|^{2}=\sum_{n\ne 0}\dfrac{1}{n^{2}}|\widehat{f'}(n)|^{2}\leq\sum|\widehat{f'}(n)|^{2}$.