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Inequality of a convex function

Let f:[0, 2] \rightarrow \mathbb{R} be a twice differentiable function such that f''(x)>0 forall x \in [0, 2]. Prove that:

    \[\int_0^2 f(x) \, \mathrm{d}x > 2 f(1)\]

Solution

The equation of the tangent of \mathcal{C}_f at x_0=1 is

    \[\left ( \varepsilon \right ): y - f(1) = f'(1) \left ( x-1 \right ) \Leftrightarrow y = f'(x) \left ( x-1 \right ) + f(1)\]

Since f''(x)>0 forall x \in [0, 2] we deduce that f is strictly convex. Thus, the tangent lies below the graph of f except at x_0=1. Hence,

    \begin{align*} f(x) \geq f'(1)\left ( x-1 \right ) + f(1) &\Rightarrow \int_{0}^{2} f(x) \, \mathrm{d}x > \int_{0}^{2} f'(1) \left ( x-1 \right ) +f(1) \int_{0}^{2} \, \mathrm{d}x \\ &\Rightarrow \int_{0}^{2} f(x) \, \mathrm{d}x > \cancelto{0}{f'(1) \left [ \frac{x^2}{2} - x \right ]_0^2} + 2f(1) \\ &\Rightarrow \int_{0}^{2} f(x) \, \mathrm{d}x > 2f(1) \end{align*}

and the conclusion follows.

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