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Inequality

Let \( f: [0,\infty] \rightarrow \mathbb R\) be a non-decreasing continuous function. Show then that the inequality.

\[ \large (z-x) \int_y^z f(u) \, du \geq (z-y) \int_x^z f(u) \, du \]

holds for any \(0\leq x\leq y \leq z \).

Note by Pratik Roy
1 year, 8 months ago

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Define \( g(x) \) as \( \dfrac{\int\limits_x ^z f(u)du}{z - x} \).
Show that \( g(x) \) is increasing - and you're done.

Indeed, \( g'(x) = \dfrac{\int\limits_x ^z f(u)du - f(x)(z - x)}{(z - x)^2}\)

Now see that \( f(u) \geq f(x)\) \( \forall \) \( u \in [x, z] \).

This means, \( \int\limits_x ^z f(u)du \geq \int\limits_x ^z f(x)du = f(x)(z - x) \) and \(g'(x) \geq 0 \)

Ameya Daigavane - 1 year, 8 months ago

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thanks sir sir what should be the approach while solving these types of problems.??

Pratik Roy - 1 year, 8 months ago

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Well, I noticed that the statement could be written in terms of a new function, \( g(x) \), because of the symmetrical way the statement is written.
If \( g(x) \) was increasing, then we would have what we wanted - and I showed that.

Ameya Daigavane - 1 year, 8 months ago

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@Ameya Daigavane sir i am looking forward to learn about countability and infinite sets . can you please suggest from where to study.

Pratik Roy - 1 year, 8 months ago

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@Ameya Daigavane thanks sir

Pratik Roy - 1 year, 8 months ago

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