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 $0\leq x\leq y \leq z$ .

#Calculus

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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

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 - 5 years, 2 months ago

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

Pratik Roy - 5 years, 2 months ago

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 - 5 years, 2 months ago

@Ameya Daigavane – thanks sir

Pratik Roy - 5 years, 2 months ago

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

Pratik Roy - 5 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$