Show that $\int_0^1 h(t)dt\geq\left(\int_0^1 f(t)dt \right)^a\left(\int_0^1 g(t)dt \right)^{1-a}$

This is the Prékopa–Leindler inequality (see e.g. here). Also here

Edit: The proof that can be found in the paper of Gardner, in the setting of the OP, is at the level of a Calculus class. Let us consider the functions $$ F(u) := \int_0^u f(x)\, dx, \qquad G(u) := \int_0^u g(x)\, dx, \qquad u \in [0,1]. $$ Since $f,g$ are continuous and strictly positive functions, we have that $F$ and $G$ are continuously differentiable and strictly increasing, with $F' = f$ and $G' = g$. If we set $F_1 := F(1)$, $G_1 := G(1)$, then $F$ is a bijection from $[0,1]$ to $[0, F_1]$ and $G$ is a bijection from $[0,1]$ to $[0, G_1]$. Let $u,v\colon [0,1] \to [0,1]$ be the functions defined by $$ u(t) := F^{-1}(F_1\, t), \qquad v(t) := G^{-1}(G_1\, t), \qquad t\in [0,1]. $$ Since $u'(t) = F_1 / f(u(t))$ and $v'(t) = G_1 / g(v(t))$ for every $t\in [0,1]$, using the AM-GM inequality we deduce that $$ w'(t) := a\, u'(t) + (1-a)\, v'(t) \geq [u'(t)]^a [v'(t)]^{1-a} = \frac{F_1^a}{f(u(t))^a}\cdot \frac{G_1^{1-a}}{g(v(t))^{1-a}} $$ and finally $$ \int_0^1 h(x)\, dx = \int_0^1 h(w(t))\, w'(t)\, dt \geq \int_0^1 f(u(t))^a g(v(t))^{1-a} \frac{F_1^a}{f(u(t))^a}\cdot \frac{G_1^{1-a}}{g(v(t))^{1-a}}\, dt = F_1^a G_1^{1-a}. $$

Now that Rigel handed us the Excalibur, I just would like to show how to derive the result for our settings.

As we will see, (a) there is nothing particular about the choice of the interval $[0,1]$. The result holds if the condition holds in an interval $I$, bounded or unbounded; (b) continuity is not crucial here. However we assume that $f$ and $g$ are measurable functions such that $f^{-1}(U)+g^{-1}(U)$ is measurable whenever $U$ is a Borel subset of $\mathbb{R}$. Certainly this is satisfied when $f$ and $g$ are continuous, and more generally when $f$ and $g$ are Borel measurable (in this case $f^{-1}(U)+g^{-1}(U)$ is universally measurable).

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Some notation first. For any function $\Phi:[0,1]\rightarrow[0,\infty)$ and $t\geq0$, we denote $\{\Phi>t\}=\{x\in[0,1]: \Phi(x)>t\}$; for any set $A\subset\mathbb{R}$, and $b\in\mathbb{R}$, $bA=\{ba:a\in A\}$; For any sets $A,B\subset\mathbb{R}$, $A+B=\{a+b:a\in A, b\in B\}$.

• The first claim is that $$a \{f>t\} +(1-a)\{g>t\}\subset\{h>t\}\tag{1}\label{one}$$ To check this, notice that if $u$ belongs to the set in the left-hand side of $\eqref{one}$ then $u=ax+(1-a)y$ for some $x\in \{f>t\}$ and $y\in\{g>t\}$ and so, $h(u)\geq f(x)^ag^{1-a}(y)>t^a t^{1-a}=t$.

• By Brunn–Minkowski's inequality and the homogeneity of Lebesgue's measure $\lambda$ (on $\mathbb{R}$) \begin{aligned} \lambda(\{h>t\})&\geq \lambda\big(a\{f>t\} +(1-a)\{g>t\}\big)\\ &\geq\lambda(a\{f>t\})+\lambda((1-a)\{g>t\})\\ &=a\lambda(\{f>t\})+(1-a)\lambda(\{g>t\}) \end{aligned}

• Fubuni's theorem leads to \begin{aligned} \int^1_0h(s)\,ds&=\int^\infty_0\lambda(\{h>t\})\,dt\\ &\geq a\int^\infty_0\lambda(\{f>t\})\,dt + (1-a) \int^\infty_0\lambda(\{g>t\})\,dt \\ &= a\int^1_0f(s)\,ds + (1-a)\int^1_0 g(s)\,ds \end{aligned}

• Finally, by the arithmetic--geometric inequality ($a\,\alpha+ (1-a)\,\beta\geq \alpha^a\beta^{1-a}$)

\begin{aligned} \int^1_0h(s)\,ds\geq a\int^1_0f(s)\,ds + (1-a)\int^1_0 g(s)\,ds \geq\Big(\int^1_0 f(s)\,ds\Big)^a\big(\int^1_0 g(s)\,ds\Big)^{1-a} \end{aligned}

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Edit: We have used a big result, namely the Brunn-Minksowki inequality. In the real line however, this result may be proven without much effort. Here is a short proof.

Brunn-Minkowski's inequality on $\mathbb{R}$. Suppose $A,B\subset\mathbb{R}$ are measurable subsets such that $A+B$ is measurable. Then $$\lambda(A+B)\geq \lambda(A) +\lambda(B)\tag{2}\label{two}$$

Proof: It suffices to assume that both $\lambda(A)$ and $\lambda(B)$ are finite. Due to the inner regularity of Lebesgue's measure, it is enough to show that $\eqref{two}$ holds for $A$ and $B$ compact. If $a^*:=\sup A$ and $b_*:=\inf B$ then $$ A+B\subset (a^*+B)\cup(A+ b_*)\supset\{a^*+b_*\}$$ If $x\in (a^*+B)\cap(A+ b_*)$ then for some $(a,b)\in A\times B$, $x=a^*+b=b_*+a$. Since $0\leq a^*-a=b_*-b\leq0$, it follows that $a=a^*$ and $b=b_*$ and so, $x=a^*+b_*$. Thus \begin{aligned} \lambda(A+B)&\geq \lambda\big((a^*+B)\cup(A+b_*)\big)\\ &=\lambda((a^*+B)+\lambda(A+b_*)-\lambda(\{a^*+b^*\})=\lambda(A)+\lambda(B) \end{aligned}

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Notes:

• The result can be generalized to higher dimensions starting from the one dimensional case and proceeding by induction with the help of Fubini's theorem.
• The higher dimensional versions of Brunn-Minkowski's inequality can be obtained form the high dimensional version of Prékopa–Leindler's theorem.

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Thanks to Rigel for the reminder of the aforementioned result which in turn reminded me of the Brunn–Minkowski inequality.

I still would like to know how rulergraham's teacher proved the statement in his Calculus class. A much simpler (but wickedly tricky) argument perhaps?