Sufficient bound on variance for an inequality to hold Let $X$ be a continuous, unimodal, symmetrical random variable on $[0,1]$ with $EX=\frac{1}{2}$ and $\operatorname{Var} (X)=\sigma^2$. Let $f$ be $X$'s density, and $F$ - the corresponding cdf. I assume that $f(\frac{1}{2})$ is large, i.e. $f(\frac{1}{2})>\frac{1}{1-2a}$, where $0<a<\frac{1}{2}$ is some given parameter and $f(0)=f(1)=0$. I would also assume $f$ is increasing on $(0,1/2)$ [edit] and $f$ differentiable.
Assume $\bar{x}<1/2$ satisfies $\bar{x}=a+(1-2a)F(\bar{x})$ for same $a$ as above (notice that $\frac{1}{2}$ also satisfies this equation, but I am interested in the smaller solution).
I would like to determine under what conditions the following inequality would hold:
$$\operatorname{Var}(X)\leq \left (\frac{1}{2}-\bar{x}\right )^2$$
Notice that both sides depend on the variability of $x$. In particular, as $\operatorname{Var}(X)$ decreases and $X$ tends to Dirac's delta $\delta_{\frac{1}{2}}$, the LHS becomes small and the RHS becomes large, i.e. tends to $(\frac{1}{2}-a)^2.$ My intuition is that if $\operatorname{Var}(X)$ is "sufficiently bounded", the inequality would hold. However, I find it difficult to formalize what bound would help me.
[edit] My best intuition so far is to have a condition $\int_0^1 xF(x)<\epsilon$, but I do not know how to use it to say anything interesting about $\bar{x}$ (of course, we want to make $\bar{x}$ "small enough").
 A: Ok, so I am going to reason why a $\int_a^{1-a}f(x)\geq 0.5+a$ type restriction is a useful type of restriction but apprently is is still not sufficient since I couldn't get rid of $F(\bar{x})$ . I'll use $\int_a^{1-a}f(x)\geq 0.5+a$ in this example because that is what I originally said but you can make make that 0.5 higher to get stronger bounds, notice however that than a also needs to be defined on an interval [0,1-b] and is no longer defined on [0,0.5] so for this a this is the strongest bound of this kind.
 It is hard to actually make this restriction work in general in terms of $\bar{x}$ I did derive an upperbound for the variance based on $a$ though which can be tranlsted into a bound in terms of $\bar{x}$ and $F(\bar{x})$ but I couldn't remove $F(\bar{x})$ and I am not aware of any way of doing that because it depends on both $a$ and $F(\bar{x})$ and without a second equation with all of them involved you can't get rid of both. Hopefully this does help though
So let's first start with that no always incrasing distribution can have a variance larger than the uniform distribution on the relevant interval. This takes care of the border case where $a$ goes to zero since $\operatorname{Var}(X)\leq 1/12 \leq \left (\frac{1}{2}-0\right )^2$.
 in fact we can use this further so let's assume x is not so close to zero than we can start to use our new restriction. First we use the definition of the variance $\operatorname{Var}(x)= \int (x-\mu)^2 f(x)dx$ now we can make that $\operatorname{Var}(x)= \int^{1}_{0} (x-\mu)^2 f(x)dx$ because we know the bounds of the domain. Then we can split that integral up into mutliple regions  so that $\operatorname{Var}(x)= \int^{a}_{0} (x-\mu)^2 f(x)dx+\int^{1-a}_{a} (x-\mu)^2 f(x)dx+\int^{1}_{1-a} (x-\mu)^2 f(x)dx$ because of symmetry we have $\operatorname{Var}(x)= 2\int^{a}_{0} (x-\mu)^2 f(x)dx+\int^{1-a}_{a} (x-\mu)^2 f(x)dx$
Now we can  calculate upperbounds over these terms. we can not calculate the size of those integrals exactly because we do not know f(x).
However since we know some things from our restriction we can derive an upperbound for each term.
Let's first note that all multiplicative terms are nonnegative so $(x-\mu)^2, f(x)\geq0$  this is important becauswe can say that given a positive function f(x) times another positive function g(x) that is smaller than 1 for all relevant values: $\int^{a}_{0} g(x) f(x) \leq \int^{a}_{0} f(x)$ so we know 
$\operatorname{Var}(x)= 2\int^{a}_{0} (x-\mu)^2 f(x)dx+\int^{1-a}_{a} (x-\mu)^2 f(x)dx \leq 2\int^{a}_{0}  f(x)dx+\int^{1-a}_{a} f(x)dx=1$ Now this gives as an upper bound but this is not really useful. But we can do better you say we can say that  $\int^{a}_{0} g(x) f(x)dx\leq\int^{a}_{0} \max_{x}(g(x)) f(x) dx\leq \int^{a}_{0} f(x)dx$.
so now we have a value we can evaluate that we know that $\int_a^{1-a}f(x)=0.5+a+b\geq 0.5+a$  with $b \geq 0$ a slack variable so we know that $\int^{1-a}_{a} (x-\mu)^2 f(x)dx \leq \max_{a\leq x\leq 1-a}(x-\mu)^2(0.5+a+b)$ and $2\int^{a}_{0} (x-\mu)^2 f(x)dx \leq 2 \max_{0\leq x\leq a}((x-\mu)^2)(1-(0.5+a+b))$ we can also easlily see that in the first the maximum expresiion evaluates to $(a-\mu)^2$ and in the second to $(0-\mu)^2$
so finally we can say $Var(x)\leq(a-\mu)^2(0.5+a+b)+2(\mu)^2(1-0.5-a-b)$
filling in $\mu$ we have a inequality for the variance
$Var(x)\leq(a-0.5)^2(0.5+a+b)+0.5(0.5-a-b)$ clearly the right hand side obtains it's maximum when b=0 because the negative coefficient for b in the second term is always larger than the positive coefficient in the first term therefore.
$Var(x)\leq(a-0.5)^2(0.5+a)+0.5(0.5-a)=a^3 - 0.5 a^2 - 0.75 a + 0.375$
Then we can fill in $a$ if we want $a=\dfrac{\bar{x}-F(\bar{x})}{1-2F(\bar{x})}$
