Showing $\frac{f(a+2b)+f(a-2b)}{2f(a)}\leq\sqrt{1-\left(\frac{b}{a(1-a)}\right)^2}$ for $f(x)=\sqrt{x(1-x)}$ with some constraints

Question

My question is from a inequality that is not proved (it is just implicitly mentioned I guess) in a book.

Specifically, let $a \in (0,1)$ and $b \in (0,1)$ with $a - 2b \geq 0$ and $a + 2b \leq 1$. Then with $f(x) := \sqrt{x(1-x)}$ we are asked to show that $$\frac{f(a+2b)+f(a-2b)}{2f(a)} \leq \sqrt{1-\left(\frac{b}{a(1-a)}\right)^2}$$

I am really stuck at this, the only related inequality that comes to mind is that $$\frac{\sqrt{1+2\alpha}+\sqrt{1-2\alpha}}{2} \leq \sqrt{1-\alpha^2},\quad \forall \alpha \in [0,\tfrac{1}{2}]$$
But I still cannot prove the desired inequality.

Any help is greatly appreciated!

Answer 1

I will just bash it. Let's work on the square of the LHS first. You will actually find that:

$$f^2(a+2b) + f^2(a-2b) = 2a(1-a) - 8b^2 = 2f^2(a) - 8b^2$$ and $$f(a+2b)f(a-2b) = \sqrt{(a^2-4b^2)((1-a)^2 - 4b^2)} = \sqrt{(f^2(a)+4b^2)^2 - 4b^2}$$ Therefore, the squared version of your inequality is:

$$LHS^2 = \dfrac{f^2(a) - 4b^2 + \sqrt{(f^2(a)+4b^2)^2- 4b^2}}{2f^2(a)}\leq 1 - \dfrac{b^2}{f^4(a)}=RHS^2$$

Now it looks like some nice substitution will simplify things:

$$\begin{cases} f^2(a) = a(1-a) = x\in(0,\frac 14) \\ 4b^2 = y\in (0,\frac 14) \end{cases}.$$ Then, you have: $$\dfrac{x-y+\sqrt{(x+y)^2-y}}{2x}\leq1 - \dfrac{y}{4x^2}\iff \dfrac{(x+y)^2-y}{4x^2}\leq \left(\frac 12-\frac{y}{4x^2}+\frac{y}{2x}\right)^2$$ and after a couple line of simplifying, you will get: $$y(1-x)(yx-y+4x^2+2xy) + 4x^2y\geq 0$$ which is trivially true. There is no equality attained in the domain you specified, but $b =y= 0$ gives an equality.

P.S: I did try an naive AM-GM in the beginning, but that resulted in $a(1-a)\geq\frac 14$, which is only true when $a=\frac 12$, so some cute AM-GM is probably not gonna cut it.