Reverse Cauchy Schwarz for integrals 
Let $f,g$ be two continuous positive functions over $[a,b]$
Let $m_1$ and $M_1$ be the minimum and maximum of $f$
Let $m_2$ and $M_2$ be the minimum and maximum of $g$
Prove that $$\sqrt{\int_a^bf^2 \int_a^b g^2}\leq \frac{1}{2}\left(\sqrt{\frac{M_1M_2}{m_1m_2}}+\sqrt{\frac{m_1m_2}{M_1M_2}}\right)\int_a^bfg$$

I can't make any significant progress on this one...
Thanks for any hint.
 A: By scaling, which does not change the ratio of maximum to minimum, we can assume
$$
\int_a^bf(x)^2\,\mathrm{d}x=\int_a^bg(x)^2\,\mathrm{d}x=1\tag{1}
$$
Trivially, we have
$$
\int_a^bf(x)g(x)\frac{f(x)}{g(x)}\,\mathrm{d}x=\int_a^bf(x)^2\,\mathrm{d}x=1\tag{2}
$$
and
$$
\int_a^bf(x)g(x)\frac{g(x)}{f(x)}\,\mathrm{d}x=\int_a^bg(x)^2\,\mathrm{d}x=1\tag{3}
$$
Since each term is non-negative,
$$
\left(\frac{M_f}{m_g}-\frac{f(x)}{g(x)}\right)\left(\frac{M_g}{m_f}-\frac{g(x)}{f(x)}\right)\ge0\tag{4}
$$
where $M_f$ and $m_f$ are the maxima and minima of $f$ and $M_g$ and $m_g$ are the maxima and minima of $g$.
Integrating $(4)$ against $f(x)g(x)$ using $(2)$ and $(3)$ gives
$$
\begin{align}
\left(\frac{M_f}{m_g}\frac{M_g}{m_f}+1\right)\int_a^bf(x)g(x)\,\mathrm{d}x
&\ge\frac{M_f}{m_g}+\frac{M_g}{m_f}\\
&\ge2\sqrt{\frac{M_f}{m_g}\frac{M_g}{m_f}}\tag{5}
\end{align}
$$
Therefore,
$$
\frac12\left(\sqrt{\frac{M_f}{m_g}\frac{M_g}{m_f}}+\sqrt{\frac{m_g}{M_f}\frac{m_f}{M_g}}\right)\int_a^bf(x)g(x)\,\mathrm{d}x\ge1\tag{6}
$$
Undoing the scaling used to get $(1)$, we get
$$
\hspace{-1cm}\frac12\left(\sqrt{\frac{M_f}{m_g}\frac{M_g}{m_f}}+\sqrt{\frac{m_g}{M_f}\frac{m_f}{M_g}}\right)\int_a^bf(x)g(x)\,\mathrm{d}x\ge\sqrt{\int_a^bf(x)^2\,\mathrm{d}x\int_a^bg(x)^2\,\mathrm{d}x}\tag{7}
$$
A: We use continuity of $f$ and $g$, to prove that the integral inequality: $$\displaystyle \int_a^b f^2\int_a^bg^2 \le \dfrac{1}{4}\left(\sqrt{\dfrac{M_1M_2}{m_1m_2}}+\sqrt{\dfrac{m_1m_2}{M_1M_2}}\right)^2\left(\int_a^bfg\right)^2$$
follows from proving the discrete case of the inequality, taking $a_k = f(a+\frac{b-a}{n}k)$ and $b_k = g(a+\frac{b-a}{n}k)$, for $k=1,\cdots,n$.
$$\sum_{k=1}^na_k^2\sum_{k=1}^nb_k^2 \le \left(\dfrac{M+m}{2\sqrt{Mm}}\sum_{k=1}^n a_kb_k\right)^2$$
where, $\displaystyle M = {\dfrac{M_1}{m_2}} \ge \dfrac{a_k}{b_k} \ge {\dfrac{m_1}{M_2}} = m$, for all $1 \le k \le n$.
Using, $\displaystyle \left(\dfrac{a_k}{b_k} - M\right)\left(\dfrac{a_k}{b_k} - m\right) \le 0 \implies a_k^2+Mmb_k^2 \le (M+m)a_kb_k$
Thus, $$\sum_{k=1}^na_k^2+Mm\sum_{k=1}^nb_k^2 \le (M+m)\sum_{k=1}^na_kb_k$$
Using Am-Gm inequality,
$$2\left(\sum_{k=1}^na_k^2\right)^{1/2}\left(Mm\sum_{k=1}^nb_k^2\right)^{1/2} \le \sum_{k=1}^na_k^2+Mm\sum_{k=1}^nb_k^2 \le (M+m)\sum_{k=1}^na_kb_k$$
Dividing both sides by $n$ and taking limit as $n \to \infty$,
$$\lim\limits_{n\to \infty}\left(\dfrac{1}{n}\sum_{k=1}^na_k^2\right)^{1/2}\left(\dfrac{1}{n}\sum_{k=1}^nb_k^2\right)^{1/2} \le \lim\limits_{n\to \infty}\dfrac{M+m}{2\sqrt{Mm}}\dfrac{1}{n}\sum_{k=1}^na_kb_k$$
$$\implies \displaystyle \sqrt{\int_a^b f^2\int_a^bg^2} \le \dfrac{1}{2}\left(\sqrt{\dfrac{M_1M_2}{m_1m_2}}+\sqrt{\dfrac{m_1m_2}{M_1M_2}}\right)\int_a^bfg$$
A: Here is a proof of a reverse Hölder inequality proven in a manner very similar to the proof of the reverse Cauchy-Schwarz inequality in my other answer. In what follows, $p,q\gt1$ and $\frac1p+\frac1q=1$. The result requested in the question is simply the case $p=q=2$.

By scaling, which does not change the ratio of maximum to minimum, we can assume
$$
\int_a^bf(x)^p\,\mathrm{d}x=\int_a^bg(x)^q\,\mathrm{d}x=1\tag{1}
$$
Trivially, we have
$$
\int_a^bf(x)g(x)\frac{f(x)^{p-1}}{g(x)}\,\mathrm{d}x=\int_a^bf(x)^p\,\mathrm{d}x=1\tag{2}
$$
and
$$
\int_a^bf(x)g(x)\frac{g(x)^{q-1}}{f(x)}\,\mathrm{d}x=\int_a^bg(x)^q\,\mathrm{d}x=1\tag{3}
$$
Since each term is non-negative,
$$
\left(\frac{M_f^{p-1}}{m_g}-\frac{f(x)^{p-1}}{g(x)}\right)\left(\frac{M_g^{q-1}}{m_f}-\frac{g(x)^{q-1}}{f(x)}\right)\ge0\tag{4}
$$
where $M_f$ and $M_g$ are the suprema and $m_f$ and $m_g$ are the infima of $f$ and $g$.
Integrating $(4)$ against $f(x)g(x)$ using $(2)$ and $(3)$ gives
$$
\begin{align}
&\left(\frac{M_f^{p-1}}{m_g}\frac{M_g^{q-1}}{m_f}+M_f^{p-2}M_g^{q-2}\right)\int_a^bf(x)g(x)\,\mathrm{d}x\\
&\ge\frac{M_f^{p-1}}{m_g}\frac{M_g^{q-1}}{m_f}\int_a^bf(x)g(x)\,\mathrm{d}x
+\int_a^bf(x)^{p-1}g(x)^{q-1}\,\mathrm{d}x\tag{5}\\
&\ge\frac{M_f^{p-1}}{m_g}+\frac{M_g^{q-1}}{m_f}\tag{6}\\
&\ge\left(q\frac{M_f^{p-1}}{m_g}\right)^{\frac1q}\left(p\frac{M_g^{q-1}}{m_f}\right)^{\frac1p}\tag{7}
\end{align}
$$
Explanation:
$(5)$: $M_f^{p-2}f(x)\ge f(x)^{p-1}$ and $M_g^{q-2}g(x)\ge g(x)^{q-1}$
$(6)$: Integrate $(4)$ against $f(x)g(x)$ and use $(2)$ and $(3)$
$(7)$: Young's Inequality
Therefore,
$$
\frac1{p^{\frac1p}q^{\frac1q}}\left[\left(\frac{M_f}{m_f}\right)^{\frac1q}\left(\frac{M_g}{m_g}\right)^{\frac1p}+\left(\frac{m_f}{M_f}\right)^{\frac1p}\left(\frac{m_g}{M_g}\right)^{\frac1q}\right]\int_a^bf(x)g(x)\,\mathrm{d}x\ge1\tag{8}
$$
Undoing the scaling used to get $(1)$, we get
$$
\begin{align}
&\frac1{p^{\frac1p}q^{\frac1q}}\left(\frac{M_f}{m_f}\right)^{\frac{p}{p+q}}\left(\frac{M_g}{m_g}\right)^{\frac{q}{p+q}}\left(1+\frac{m_f}{M_f}\frac{m_g}{M_g}\right)\int_a^bf(x)g(x)\,\mathrm{d}x\\
&\ge\left(\int_a^bf(x)^p\,\mathrm{d}x\right)^{\frac1p}\left(\int_a^bg(x)^q\,\mathrm{d}x\right)^{\frac1q}\tag{9}
\end{align}
$$
