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 \(0\leq x\leq y \leq z \).

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 \(0\leq x\leq y \leq z \).

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## Comments

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TopNewestDefine \( g(x) \) 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 ago

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– Pratik Roy · 1 year ago

thanks sir sir what should be the approach while solving these types of problems.??Log in to reply

If \( g(x) \) was increasing, then we would have what we wanted - and I showed that. – Ameya Daigavane · 1 year ago

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– Pratik Roy · 1 year ago

sir i am looking forward to learn about countability and infinite sets . can you please suggest from where to study.Log in to reply

– Pratik Roy · 1 year ago

thanks sirLog in to reply