MHB How does Jensen's inequality apply to this problem?

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The discussion focuses on applying Jensen's inequality to demonstrate that the exponential of the integral of a function is less than or equal to the integral of the exponential of that function. The problem states that for an integrable function f over the interval [0,1], the inequality exp[integral of f] ≤ integral of exp(f) holds. The solution highlights that the exponential function is convex, which is crucial for applying Jensen's inequality. By integrating the derived inequalities, the conclusion follows that the initial inequality is satisfied. This illustrates a specific application of Jensen's inequality in the context of integrable functions.
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Here's this week's problem!

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Problem
: Let $f$ be integrable over $[0,1]$. Show that
\[\exp\left[\int_0^1 f(x)\,dx\right] \leq \int_0^1\exp(f(x))\,dx.\]

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This week's problem was correctly answered by Ackbach and Opalg. You can find Opalg's solution below.

[sp]This is a particular case of Jensen's inequality.

The exponential function is convex, so the tangent at any point lies below the curve (except at the point where they touch). The equation of the tangent at the point $(t,e^t)$ is $y = e^tx + (1-t)e^t.$ It follows that $e^x \geqslant e^tx + (1-t)e^t. \quad(*)$

Let $$J = \int_0^1 \!\!f(x)\,dx.$$ Take $t=J$ in (*) (and replace $x$ by $f(x)$) to see that $ \exp(f(x)) \geqslant e^Jf(x) + (1-J)e^J.$

Now integrate that from $0$ to $1$: $$\int_0^1\!\! \exp(f(x))\,dx \geqslant e^J\!\!\int_0^1\!\!f(x)\,dx + \int_0^1\!\!(1-J)e^J\,dx = e^JJ + (1-J)e^J = e^J = \exp\left[ \int_0^1 \!\!f(x)\,dx\right].$$[/sp]