Solution prove an integral-inequality with certain initial condition Following the title, my problem is
With $f$ is continuous in $[0;1]$ and $\int\limits_0^1 {xf\left( x \right)dx}  = 0$, prove
$\int\limits_0^1 {{{\left| {f\left( x \right)} \right|}^2}} dx \ge 4{\left( {\int\limits_0^1 
{f\left( x \right)dx} } \right)^2}$
Solution:
We have
$$\int\limits_0^1 {\left( {3x - 2} \right)f\left( x \right)dx}  = 3\int\limits_0^1 {xf\left( x \right)dx}  - 2\int\limits_0^1 {f\left( x \right)dx}  =  - 2\int\limits_0^1 {f\left( x \right)dx} $$
Thus
$$4{\left( {\int\limits_0^1 {f\left( x \right)dx} } \right)^2} = {\left( {\int\limits_0^1 {\left( {3x - 2} \right)f\left( x \right)dx} } \right)^2} \le \int\limits_0^1 {{{\left( {3x - 2} \right)}^2}dx} \int\limits_0^1 {f{{\left( x \right)}^2}dx}  = \int\limits_0^1 {f{{\left( x \right)}^2}dx}$$
because $\int\limits_0^1 {{{\left( {3x - 2} \right)}^2}dx}  = 1$
Can anyone explain why we know to multiple $(3x-2)$ to integral to have that solution, for some  similar problem we also multiple a polynomial to integral, how can find that polynomial?
 A: Let $g(x):=ax+b$ for some real constants $a\,(\neq 0)$ and $b$
Then, $$\int_0^1 g(x)f(x)\,\mathrm dx=a\underbrace{\int_0^1xf(x)}_{=\,0}+b\int_0^1f(x)\,\mathrm dx=b\int_0^1f(x)\,\mathrm dx$$
Now, square both sides of the above and use Cauchy-Schwarz,
$$b^2\left(\int_0^1f(x)\,\mathrm dx\right)^2=\left(\int_0^1g(x)f(x)\,\mathrm dx\right)^2\le\left(\int_0^1 g(x)^2\,\mathrm dx\right)\left(\int_0^1f(x)^2\,\mathrm dx\right)$$
As noted in the comments, $\int_0^1 g(x)^2\,\mathrm dx=\int_0^1(ax+b)^2\,\mathrm dx=\frac 13(a^2+3b(a+b))$ and this value is $\geq 0$ because $g(x)^2\geq 0$, so we can divide by it without flipping the inequality sign.
Using this with the above, our result becomes,
$$\int_0^1 f(x)^2\,\mathrm dx=\int_0^1 |f(x)|^2\,\mathrm dx\geq \frac{b^2}{\frac 13(a^2+3b(a+b))}\left(\int_0^1f(x)\,\mathrm dx\right)^2$$
for any real constants $a,b$ where $a\neq 0$
Denote $C(a,b):=\dfrac{b^2}{\frac 13(a^2+3b(a+b))}$; choose constants $a,b$ suitably such that $C(a,b)=4$ to show the desired result.
This can be done by choosing $a,b$ such that the denominator of $C$, which is simply $\int_0^1 g(x)^2\,\mathrm dx=\int_0^1(ax+b)^2\,\mathrm dx$, equals $1$ and the numerator $b^2$ equals $4$, which, as already stated in the comments, can be done only with $(a,b)=(3,-2)$ or $(-3,2)$

Bonus exercise: Show that for any real $a,b$ ($a\ne 0$), we have $C(a,b)\le 4$, ie, the inequality in the desired result cannot be refined any further than $\ge 4\left(\int_0^1 f(x)\,\mathrm dx\right)^2$
Hint: $(2a+3b)^2\geq 0$ for all reals $a,b$
