Suppose $0 < r \leq a(t) \leq R < \infty$ is a continuous function on $[0, 1]$ and consider $$w(t) = \frac{1}{a(t)} \left( \int _0 ^1 \frac{1}{a(s)} ds \right)^{-1}.$$
Are there any attainable upper bounds that can be put on the following quantities: $$\int_0^1 w^2 (x) dx, \quad \int_0^1 w_1 (x) w_2 (x) dx $$ where $w_1(x), w_2(x)$ are two functions of the above form for different $a_i (t)$ with the above bound? Specifically, are there any bounds in terms of $r, R$, which can actually be achieved? If not, why?
Here are some of my thoughts. Evidently, $$\int_0^1 w(t) dt = 1 $$ and moreover, from the definition of $a(t)$, $$\frac{r}{R} \leq w(t) \leq \frac{R}{r}. $$ These bounds could be used to bound the two quantities of interest, but the problem is these bounds on $w(t)$ are not actually attainable since they require $a(t)$ to be either $r$ or $R$ on all of $[0, 1]$, which cannot happen unless $r=R$, which is a sort of trivial case. Additionally, I know that the largest value of a function can be bounded by the average and integral of its derivative, i.e. this estimate, but reasoning from this doesn't give an upper bound, and moreover, we're only given that $a(t)$ is continuous.
Is there any bound in terms of $r$ and $R$ which can be attained?
The Cauchy-Schwarz inequality implies $$ \int_0^1 w_1(t)w_2(t)dt \leq \sqrt{\int_0^1w_1(t)^2dt}\sqrt{\int_0^1 w_2(t)^2dt}$$ and this bound can be achieved by using $w_1=w_2$. It suffices to find a function that maximizes $\int_0^1 w(t)^2dt$.
Fix real numbers $r,R$ with $0<r\leq R$. For a measurable function $a:[0,1]\rightarrow[r,R]$ define $$w(t) = \frac{1/a(t)}{\int_0^11/a(s)ds} \quad \forall t \in [0,1]$$
Claim: For any measurable function $a:[0,1]\rightarrow [r,R]$ we have $$ 1\leq \int_0^1 w(t)^2dt \leq \sup_{p\in[0,1]} \left\{\frac{\frac{p}{r^2} + \frac{(1-p)}{R^2}}{\left(\frac{p}{r} + \frac{1-p}{R} \right)^2} \right\}$$ The lower bound is achieved by $a(t)=r$ for all $t \in [0,1]$. The upper-bound is achieved by $$ a^*(t) =\left\{\begin{array}{cc} r & \mbox{ if $t \in [0,p^*]$} \\ R & \mbox{ else} \end{array}\right.$$ where $p^*$ is the maximizer of the upper-bound expression. We can approximate $a^*(t)$ arbitrarily closely by a continuous function.
The lower bound is Jensen's inequality.
The upper bound can be argued by showing that if we fix $\int_0^1 a(t)=v$ (where $r\leq v\leq R$), then $\int_0^1w(t)^2dt$ can be maximized by a function $a(t)$ that always takes values in the 2-element set $\{r,R\}$. The 2-valued argument is kind of nice: Let $f(t)$ be any measurable function that satisfies $f_{min}\leq f(t)\leq f_{max}$ for all $t \in [0,1]$, and such that $\int_0^1f(t)dt=v$. Treat $f_{min}, f_{max}, v$ as fixed parameters. Define $m=(f_{max}+f_{min})/2$. Then $$\int_0^1(f(t)-m)^2=\left(\int_0^1f(t)^2dt\right) - 2mv + m^2$$ so maximizing $\int f(t)^2$ amounts to maximizing $\int (f(t)-m)^2$, and clearly $$(f(t)-m)^2\leq (f_{max}-f_{min})^2/4 \quad \forall t$$ which can be achieved by $f(t) \in \{f_{min}, f_{max}\}$ for all $t$, and using these 2 values we can also satisfy the constraint $\int_0^1 f(t)dt = v$ by using $f(t)=f_{min}$ for some fraction of time $p$, and $f(t)=f_{max}$ for fraction of time $1-p$.