Show that $\int_0^1 h(t)dt\geq\left(\int_0^1 f(t)dt \right)^a\left(\int_0^1 g(t)dt \right)^{1-a}$ Let $ a \in (0,1)$ and $f,g,h :[0,1]\to (0,\infty)$ be continuous functions satisfying
$h(ax+(1-a)y)\geq f(x)^ag(y)^{1-a}$ for all $x,y \in [0,1]$.
Prove that
$$ \int_0^1 h(t)dt\geq\left(\int_0^1 f(t)dt  \right)^a\left(\int_0^1 g(t)dt  \right)^{1-a} $$
I tried integrating left side with respect to $x$ and $y$ separately, but it didn't work.
Edit: Number '$a$' is chosen only once, the condition does not have to hold for values of '$a$' other then the one we chose.
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
$$
\int_0^1 h(x)\, dx = \int_0^1 h(w(t))\, w'(t)\, dt
\geq \int_0^1 f(u(t))^a g(v(t))^{1-a} \frac{F_1^a}{f(u(t))^a}\cdot \frac{G_1^{1-a}}{g(v(t))^{1-a}}\, dt
= F_1^a G_1^{1-a}.
$$
We assume here that the inequality (1) below holds for all $a \in\, ]0,1[$. And we prove that conclusion is valid for all $a \in\, ]0,1[$.
Observe that $f,g&gt;0$ on $[0,1]$ implies that, for each fixed $x,y\in [0,1]$, the
exponential functions $a\mapsto f(x)^{a}$, $a\mapsto g(x)^{1-a}$ are well-defined. Recall that the exponential functions are continuous on $\mathbb{R}$. Now, since $h$ is also continuous on $[0,1]$, then both sides of the inequality
$$h(ax+(1-a)y)\geq f(x)^{a}\,g(y)^{1-a}\qquad \qquad (1)$$
are continuous in $a$. This justifies taking the limit of both sides of (1) as $a\rightarrow 0^{+}$ and as $a\rightarrow 1^{-}$. So, we have that
$$h(x)\geq f(x),\forall x\in[0,1]\Longrightarrow
\int_{0}^{1}h(t)dt\geq \int_{0}^{1}f(t)dt\Longrightarrow
\left(\int_{0}^{1}h(t)dt\right)^{a}\geq \left(\int_{0}^{1}f(t)dt\right)^{a},\qquad \qquad (2)$$
and analogously,
$$h(x)\geq g(x),\forall x\in[0,1]\Longrightarrow
\int_{0}^{1}h(t)dt\geq \int_{0}^{1}g(t)dt\Longrightarrow
\left(\int_{0}^{1}h(t)dt\right)^{1-a}\geq \left(\int_{0}^{1}g(t)dt\right)^{1-a},\qquad \qquad (3)$$
The assertion follows from the inequalities (2) and (3).
Remark, we can not use the implication $\int_{0}^{1}\phi(t)dt\geq \int_{0}^{1}\psi(t)dt\Longrightarrow
\left(\int_{0}^{1}\phi(t)dt\right)^{\alpha}\geq \left(\int_{0}^{1}\psi(t)dt\right)^{\alpha}$
without the assumption
$\int_{0}^{1}\psi(t)dt&gt;0$, and $\alpha &gt;0$.
We here have $\int_{0}^{1}f(t)dt,\int_{0}^{1}g(t)dt&gt;0$
by the assumption that both $f,g$ are positive on $[0,1]$.