Statement 2: For every continuous $f : [0,\frac{\pi}{2}] \to (0,+\infty)$ satisfying $\int_0^{\pi/2} f(t)\sin(t)dt < f(\frac{\pi}{2})$, then $$ \int_0^{\pi/2} \sqrt{f(t)}dt < \frac{\pi}{2}\sqrt{f(\frac{\pi}{2})} $$
The Statement 2 is from this problem. The mathworker21 prove it is false by prove it is equivalent to another statement. I was disturbed by his proof (although I don't find material mistake). Besides, he did not give a direct counterexample. Therefore, I want an computable counterexample to end my doubt.
It is naturally to use a computer to construct a computable counterexample. First, define a function $f_0$ on $\left[0,\tfrac{\pi}2\right]$ by putting $$f_0(t)= \cases{1.124, & when $0\le t<0.2$,\\ 0.9, & when $0.2\le t<0.3$,\\ 1,& when $0.3\le t<\pi/2$.} $$ The calculations with Mathcad yield that the function $f_0$ satisfies the required inequalities, that is $$\int_0^{\pi/2} f_0(t)\sin(t)dt=0.9999987... <1=f_0\left(\frac{\pi}{2}\right),$$ $$\int_0^{\pi/2} \sqrt{f_0(t)}dt =1.5777...>1.5707...=\frac{\pi}{2}\sqrt{f_0\left(\frac{\pi}{2}\right)}.$$
Now observe that if $f$ is any continuous piecewise linear function from $\left[0,\tfrac{\pi}2\right]$ to $[0.9,1.124]$ which is equals to $f_0$ everywhere but sufficiently small neighborhoods of the points $0.2$ and $0.3$ then $f$ satisfies the required inequalities too.