$f, g: [0,1] \to \mathbb{R}$ are continuous functions. Show that $\text{Var}(fg) \le 2\text{Var}(f)M(g)^2 + 2\text{Var}(g)M(f)^2$ where $M(h) = \max\limits_{x \in [0,1]} |h(x)|, \text{Var}(h) = \int_0^1 (h(x)-\mu_h)^2 \, dx, \mu_h = \int_0^1 h(x) \, dx.$
How do you get started on this problem? If one of $f, g$ is constant, the result is true. I also noticed that $$M(f)^2 \ge \mu_f^2, M(f) \ge \text{Var}(f)$$ and $$\mu_f^2 \text{Var}(g) + \mu_g^2 \text{Var}(f) + \text{Var}(f)\text{Var}(g) = (\int_0^1 f^2)(\int_0^1 g^2) - \mu_f^2 \mu_g^2$$ If you assume $\mu_f = \mu_g = 0,$ the inequality reduces to $\int f^2 g^2 - \mu_{fg}^2 \le \int f^2g^2 \le 2(\int f^2)M(g)^2 + 2(\int g^2)M(f)^2,$ which I have solved below.
I have no idea how to tie all of these observations together. Does anyone have an idea about how you can make progress on this problem? I think that the Cauchy-Schwarz inequality $(\int fg)^2 \le (\int f^2)(\int g^2)$ will show up in some way but I'm not sure where's the best point to use it. Every time I see a spot where it applies, the inequality I'm investigating is pointing in the wrong direction for it to be useful.
Here's another idea: Replace $f, g$ with $fM(f), gM(g)$ so that $|f|, |g| \le 1$ now and $1$ is achieved at some point. In the case $\mu_f = \mu_g = 0,$ the inequality reduces to $\int f^2 g^2 \le 2\int (f^2+g^2),$ or $\int (2f^2 - f^2 g^2 + 2g^2) \ge 0.$ With $a=f^2, b=g^2,$ we have $2(a+b) \ge 4\sqrt{ab} \ge ab$ since $ab \le 16,$ so this is true. However, I am very suspicious about just how much freedom there is given that we actually have $ab \le 1.$ The general case will not be this easy.
Using this trick for the general case and expanding everything out gives $\int f^2 g^2 + 2\mu_f^2 + 2\mu_g^2 \le \mu_{fg}^2 + 2\int (f^2+g^2) \, \, (*)$ as the inequality to prove.
Update: I just recalled the tactic of turning squares of integrals into double integrals. We may rewrite $(*)$ as $$\int\limits_{[0,1]^2} [2f(x)^2+2g(x)^2+f(x)g(x)f(y)g(y) - 2f(x)f(y) - 2g(x)g(y) - f(x)^2g(x)^2] \, dx dy\ge 0$$ and we still have the condition that $|f|, |g| \le 1.$
If you symmetrize the integrand by replacing $2f(x)^2, 2g(x)^2, f(x)^2g(x)^2$ with $f(x)^2+f(y)^2, g(x)^2+g(y)^2, 0.5(f(x)^2g(x)^2+f(y)^2g(y)^2),$ then let $a=f(x), b=f(y), c=g(x), d=g(y),$ the integrand becomes $a^2+b^2+c^2+d^2+abcd-2ab-2cd-0.5(a^2c^2+b^2d^2) = (a-b)^2 + (c-d)^2 - 0.5(ac-bd)^2.$ Thus, we just need to show $(a-b)^2 + (c-d)^2 \ge 0.5(ac-bd)^2$ given $a,b,c,d \in [-1,1].$ Let $A = (0,0), B = u = (a, c), C = v = (b, d),$ and this becomes $BC^2 = ||u-v||^2 \ge 0.5 \det [u,v]^2 = 2 [ABC]^2,$ or $BC \ge \sqrt{2} \cdot [ABC].$ If $h$ is the length of the altitude from $A,$ then this finally reduces to $h \le \sqrt{2}.$ The altitude is shorter than the side lengths. In particular, $h \le AB = ||u|| = \sqrt{a^2+c^2} \le \sqrt{2}.$ I'll add this as an answer and look out for simpler solution.
If $M(f) = 0$ or $M(g) = 0,$ then $f=0$ or $g=0$ and the inequality trivially holds. Else, let $h_1 = f/M(f), h_2 = g/M(g)$ so that the inequality becomes $\text{Var}(h_1 h_2) \le 2\text{Var}(h_1) + 2\text{Var}(h_2).$ After expanding everything out, this becomes $$2 \int (h_1^2 + h_2^2) + \left(\int h_1 h_2\right)^2 - 2\left(\int h_1\right)^2 - 2\left(\int h_2\right)^2 - \int h_1^2 h_2^2 \ge 0.$$
We can express this as an inequality for a double integral:
$$\int\limits_{[0,1]^2} \left[ h_1(x)^2 + h_1(y)^2 + h_2(x)^2 + h_2(y)^2 + h_1(x)h_2(x)h_1(y)h_2(y) - 2h_1(x)h_1(y) - 2h_2(x)h_2(y) - 0.5(h_1(x)^2h_2(x)^2+h_1(y)^2h_2(y)^2)\right] \ge 0.$$
Let $a = h_1(x), b = h_1(y), c = h_2(x), d = h_2(y).$ The integrand is $(a-b)^2 + (c-d)^2 - 0.5(ac-bd)^2,$ so it suffices to show $(a-b)^2 + (c-d)^2 \ge 0.5(ac-bd)^2.$ Let $A = (0,0), B = (a,c), C = (b,d).$ Then the inequality becomes $BC^2 \ge 2[ABC]^2 \Leftrightarrow BC \ge \sqrt{2} [ABC].$ Letting $h$ be the altitude from $A$ to $BC,$ this reduces to $h \le \sqrt{2}.$ Since the altitude is the shortest distance to BC, we have $h \le AB = \sqrt{a^2+c^2} \le \sqrt{2}$ as desired.