i think you can solve by plugging x->x+a,x->x+2a etc.

Directly from the equality given: f(x+a) ≥ 1/2 for all x, and hence f(x) ≥ 1/2 for all x. So f(x+2a) = 1/2 + √( f(x+a) - f(x+a)2 ) = 1/2 + √f(x+a) √(1 - f(x+a)) = 1/2 + √(1/4 - f(x) + f(x)2) = 1/2 + (f(x) - 1/2) = f(x). So f is periodic with period 2a. We may take f(x) to be arbitrary in the interval [0,1). For example, let f(x) = 1 for 0 ≤ x < 1, f(x) = 1/2 for 1 ≤ x < 2. Then use f(x+2) = f(x) to define f(x) for all other values of x. kalva

Since $\sqrt{f(x)-(f(x))^{2}}\geq 0$, $f(x+a)\geq \frac{1}{2}$ and $f(x)\geq \frac{1}{2}$. Squaring the first equation, we obtain: $(f(x+a))^{2}-f(x+a)+\frac{1}{4}=f(x)-(f(x))^{2}$, which is equivalent to: $|f(x)-\frac{1}{2}|=\sqrt{f(x+a)-(f(x+a))^{2}}$. (*) Plugging $x+a$ instead of $x$ into the original equation, we get $f(x+2a)-\frac{1}{2}=\sqrt{f(x+a)-(f(x+a))^{2}}$. Backing this into (*) and applying the condition $f(x)\geq \frac{1}{2}$, we get: $ f(x)-\frac{1}{2}=f(x+2a)-\frac{1}{2} $, or equivalently $f(x)=f(x+2a)$.

The functional equation is equivalent to $f(x+a)=\sqrt{f(x)(1-f(x))}+1/2 \implies (f(x+a)-1/2)^2=1/4-(f(x)-1/2)^2 \implies f(x+2a)=f(x).$ Hence $f$ is periodic. For example, $f(x)=\frac{1}{2}(1+|\cos \frac{\pi x}{2}|),$ which is seen to work. And thus we're done.

A way of aproaching this problem is to simply substitute $x=x+a$. $\therefore f(x+2a)=\frac{1}{2}+\sqrt{f(x+a)-f(x+a)^2}$ Now substitute $ f(x+a)={1\over2}+\sqrt{f(x)-f(x)^2}$ $\therefore f(x+a)-f(x+a)^2={1\over2}+\sqrt{f(x)-f(x)^2} - ({1\over2}+\sqrt{f(x)-f(x)^2})^{2}={1\over4}-f(x)+f(x)^2=(f(x)-\frac{1}{2})^2.$ $\therefore f(x+2a)=\frac{1}{2}+\sqrt{(f(x)-\frac{1}{2})^2}=\frac{1}{2}+|f(x)-\frac{1}{2}|.$ Note that $\sqrt{a^2}=|a|\neq a$ for negative $a$. So we must prove $f(x)\geq\frac{1}{2} \forall x\in \mathbb{R}$ to complete our proof. Luckily, this is easy as $\sqrt a\geq 0 \forall a\in\mathbb{R} \Rightarrow f(x+a)\geq\frac{1}{2} \Rightarrow f(x)\geq \frac{1}{2} \forall x\in \mathbb{R}$. $\therefore f(x+2a)=f(x) \forall x\in \mathbb{R}$ and we are done! So, $f(x)$ is periodic with period $2a$. Finding an example is pretty elusive, but the periodicity should give us the hint of using trignometry. But $f(x)\geq\frac{1}{2} \forall x\in \mathbb{R}.$ and there is no such condition on trignometric functions, we may also have to use the modulus function. Combining these two along with the bound on $f(x)$ we get that the function may be of the form$f(x)=\frac{1}{2}+|g(x)|$ where $g(x)$ may be a trignometric function which contains $x$. After fiddling around with it a little bit we see that $f()=\frac{1}{2}+\frac{1}{2}|cos(\frac{\pi x}{2})|$ satisfies the condition, completing the 2nd part of the problem! Hope you liked it!!!