# 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
2 years, 4 months ago

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## Comments

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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 $$g'(x) \geq 0$$

- 2 years, 4 months ago

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

- 2 years, 4 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.

- 2 years, 4 months ago

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

- 2 years, 4 months ago

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thanks sir

- 2 years, 4 months ago

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