Inequality using Cauchy Buniakovsky Schwarz and Chebysev maybe? Can someone help me with this problem? Thank you!

Let $f : [0,1] \to [0, \infty)$, $f$ is nondecreasing and not null and $0 < a < b$. Show that: $$1-\left(\frac{a-b}{a+b+1}\right)^2 \leq \frac{\left(\int_0^1 x^{a+b}f(x) \, dx\right)^2}{\int_0^1 x^{2a}f(x) \, dx \cdot \int_0^1 x^{2b}f(x) \, dx} < 1.$$

My approach:
The right member is easy, just using Cauchy Buniakovski Schwarz and is already finished. The left member i don't know to do it. Maybe someone knows? Thank you!
 A: The middle term is invariant under multiplication of $f$ with a positive constant, hence for convenience we can assume $f(1) = 1$. The left hand side of the inequality is
$$\frac{(1+2a)(1+2b)}{(a+b+1)^2}\,,$$
thus the inequality we want to prove is
$$\int_0^1 \bigl((1+2a)x^{2a}\bigr)f(x)\,dx \int_0^1 \bigl((1+2b)x^{2b}\bigr)f(x)\,dx
\leqslant \Biggl(\int_0^1 \bigl((a+b+1)x^{a+b}\bigr)f(x)\,dx\Biggr)^2\,. \tag{1}$$
Now it is convenient to use the Riemann–Stieltjes integral. Integration by parts yields
$$\int_0^1 (c+1)x^cf(x)\,dx = x^{c+1}f(x)\biggr\rvert_0^1 - \int_0^1 x^{c+1}\,df(x) = 1 - \int_0^1 x^{c+1}\,df(x)$$
for all $c > 0$, and we have
$$0 \leqslant \int_0^1 x^{c+1}\,df(x) \leqslant f(1) - f(0) \leqslant 1\,.$$
Applying the Cauchy–Buniakovsky–Schwarz inequality to
$$\int_0^1 x^{a+b+1}\,df(x) = \int_0^1 x^{a + \frac{1}{2}} x^{b + \frac{1}{2}}\,df(x)$$
then yields
\begin{align}
\Biggl(\int_0^1 \bigl((a+b+1)x^{a+b}\bigr)f(x)\,dx\Biggr)^2
&= \Biggl(1 - \int_0^1 x^{a+b+1}\,df(x)\Biggr)^2 \\
&\geqslant \Biggl(1 - \sqrt{\int_0^1 x^{2a+1}\,df(x)}\sqrt{\int_0^1 x^{2b+1}\,df(x)}\Biggr)^2
\end{align}
and
$$(1 - uv)^2 - (1 - u^2)(1-v^2) = (u-v)^2 \geqslant 0$$
with
$$u = \sqrt{\int_0^1 x^{2a+1}\,df(x)}\qquad\text{and}\qquad v = \sqrt{\int_0^1 x^{2b+1}\,df(x)}$$
finally proves $(1)$.
If the Riemann–Stieltjes integral is not yet available, approximate $f$ by continuously differentiable nondecreasing functions $f_n \leqslant f_{n+1}$. Then the above argument can be made for the Riemann integral (replacing $df_n(x)$ with $f_n'(x)\,dx$), and since $(1)$ holds for each $f_n$ it also holds for the limit $f$.
