$3\int_{0}^{1}(f'(x))^2dx \geq (2\int_{0}^{1}f(x)dx)^2 \impliedby 2\int_{0}^{\frac{1}{2}}f(x)\,\mathrm dx=\int_{\frac{1}{2}}^{1}f(x) \,\mathrm dx$ Let $f : \mathbb{R} \to \mathbb{R} $ be a differentiable function.  Suppose that $2\int_{0}^{\frac{1}{2}}f(x)\,\mathrm dx=\int_{\frac{1}{2}}^{1}f(x) \,\mathrm dx$
Show that $$3\int_{0}^{1}(f'(x))^2 \,\mathrm dx \geq (2\int_{0}^{1}f(x)\,\mathrm dx)^2$$
 A: We argue about the function $u(t):=f'(t)$. One then has
$$f(x)=c+\int_{1/2}^x u(t)\ dt$$
for some $c\in\Bbb R$. Compute
$$\eqalign{\int_0^{1/2}f(x)\ dx&={c\over 2}-\int_0^{1/2} t u(t)\ dt={c\over 2}-a\cr 
\int_{1/2}^1 f(x)\ dx&={c\over 2}+\int_{1/2}^1(1- t) u(t)\ dt={c\over2}+b\cr}$$
with
$$a:=\int_0^{1/2} t u(t)\ dt, \qquad b:=\int_{1/2}^1(1- t) u(t)\ dt\ .$$ The condition ${c\over 2}+b=2\bigl({c\over2}-a\bigr)$  enforces $c=2b+4a$, so that we obtain
$$\int_0^1 f(x)\ dx=\biggl({c\over 2}-a\biggr)+\biggl({c\over2}+b\biggr)=3(a+b)\ ,$$
or
$$\int_0^1 f(x)\ dx= 3\int_0^1 g(t)\>u(t)\ dt\ ,\quad{\rm with}\quad g(t):=\cases{t&$(0\leq t\leq{1\over2})$\cr 1-t\quad&$({1\over2}\leq t\leq1)$\cr}\ .$$
By Schwarz' inequality
$$\int_0^1 u^2(t)\ dt\cdot\int_0^1 g^2(t)\ dt\geq \left(\int_0^1 g(t) u(t)\ dt\right)^2={1\over9}\left(\int_0^1 f(x)\ dx\right)^2\ .$$
Since
$$\int_0^1 g^2(t)\ dt=2 \int_0^{1/2} t^2\ dt={1\over12}$$
the stated inequality follows.
A: Integration by parts gives
$$
\int_0^{1/2}tf'(t)\,\mathrm{d}t=\frac12f\left(\frac12\right)-\int_0^{1/2}f(t)\,\mathrm{d}t\tag{1}
$$
and
$$
\int_{1/2}^1(1-t)f'(t)\,\mathrm{d}t=-\frac12f\left(\frac12\right)+\int_{1/2}^1f(t)\,\mathrm{d}t\tag{2}
$$
Adding $(1)$ and $(2)$ and applying the hypothesis yields
$$
\begin{align}
\int_{1/2}^1(1-t)f'(t)\,\mathrm{d}t+\int_0^{1/2}tf'(t)\,\mathrm{d}t
&=\int_{1/2}^1f(t)\,\mathrm{d}t-\int_0^{1/2}f(t)\,\mathrm{d}t\\
&=\frac23\int_0^1f(t)\,\mathrm{d}t-\frac13\int_0^1f(t)\,\mathrm{d}t\\
&=\frac13\int_0^1f(t)\,\mathrm{d}t\tag{3}
\end{align}
$$
$(3)$ and Cauchy-Schwarz say that
$$
\begin{align}
\frac13\left|\,\int_0^1f(t)\,\mathrm{d}t\,\right|
&\le\left(\int_0^{1/2}t^2\,\mathrm{d}t+\int_{1/2}^1(1-t)^2\,\mathrm{d}t\right)^{1/2}\left(\int_0^1f'(t)^2\,\mathrm{d}t\right)^{1/2}\\
&=\frac1{\sqrt{12}}\left(\int_0^1f'(t)^2\,\mathrm{d}t\right)^{1/2}\\
4\left(\int_0^1f(t)\,\mathrm{d}t\right)^2&\le3\int_0^1f'(t)^2\,\mathrm{d}t\tag{4}
\end{align}
$$
A: by integrating by parts,
$$\int_0^{\frac{1}{2}}g_1(x)f'(x)\,\mathrm dx=g_1(x)f(x)\bigg|_0^{\frac{1}{2}}-\int_0^{\frac{1}{2}}g_1'(x)f(x)\,\mathrm dx \tag{1}$$
$$\int_{\frac{1}{2}}^1g_2(x)f'(x)\,\mathrm dx=g_2(x)f(x)\bigg|_{\frac{1}{2}}^1-\int_{\frac{1}{2}}^1g_2'(x)f(x)\,\mathrm dx \tag{2}$$
Since $2\int_{0}^{\frac{1}{2}}f(x)\,\mathrm dx=\int_{\frac{1}{2}}^{1}f(x) \,\mathrm dx$, we let $g_1'(x)=a_1\neq0$ and $g_2'(x)=b_1\neq0$.
$$g(x)=\begin{cases}
g_1(x)=a_1x+a_0,x\in[0,\tfrac{1}{2})\\
g_2(x)=b_1x+b_0,x\in(\tfrac{1}{2},1]
\end{cases}$$
Let
$g_1(x)f(x)\bigg|_0^{\frac{1}{2}}+g_2(x)f(x)\bigg|_{\frac{1}{2}}^1=0$, we get
$$\begin{cases}
g_1(0)=g_2(1)=0\\
g_1(\tfrac{1}{2})=g_2(\tfrac{1}{2})
\end{cases} \Rightarrow\quad \begin{cases}
a_0=b_1+b_0=0\\
\tfrac{1}{2}a_1+a_0=\tfrac{1}{2}b_1+b_0
\end{cases}$$
Hence, we get $a_0=0,a_1=a_1,b_0=-a_1,b_1=a_1$,
$$g(x)=\begin{cases}
g_1(x)=a_1x,x\in[0,\tfrac{1}{2})\\
g_2(x)=-a_1x+a_1,x\in(\tfrac{1}{2},1]
\end{cases}$$
from $(1)+(2)$, we have
\begin{align*}
\int_0^1g(x)f'(x)\,\mathrm dx&=-\int_0^{\frac{1}{2}}g_1'(x)f(x)\,\mathrm dx-\int_{\frac{1}{2}}^1g_2'(x)f(x)\,\mathrm dx\\
&=-\frac{a}{3}\int_0^1f(x)\,\mathrm dx+\frac{2a}{3}\int_0^1f(x)\,\mathrm dx=\frac{a}{3}\int_0^1f(x)\,\mathrm dx
\end{align*}
by Cauchy-Schwarz inequality, we get
$$\left(\frac{a}{3}\int_0^1f(x)\,\mathrm dx\right)^2\leqslant \int_0^1g^2(x)\,\mathrm dx\int_0^1(f'(x))^2\,\mathrm dx$$
and
$$\int_0^1g^2(x)\,\mathrm dx=\frac{a_1^2}{24}+\frac{a_1^2}{24}=\frac{a_1^2}{12}$$
Therefore,
$$\left(\int_0^1f(x)\,\mathrm dx\right)^2\leqslant \frac{3}{4}\int_0^1(f'(x))^2\,\mathrm dx$$
