proof of inequality calculus

[rustic heath]

Is this proof correct? If not why not?

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[rustic heath]

My problem lies on wether I'm allowed to use the derivative on the given inequality. I'm kinda confused about when I can actually use the derivative in an equality. For example if I have the given inequality:
f(x)>0 then u can't say f'(x)>0
But in other cases where i have something like
f(x)+g(x)>h(x) u can use the derivative, can't you? What's the difference and how do i distinguish when i can/can't use it?

[chilly mortar]

Can you rewrite separately what the statement to prove is?

[rustic heath]

For every x>0, and with f' strictly decreasing, prove that:

2f(x)-f(2x)>4

The exercise gives us f(x)=-x⁴+2ax³-24x²+2x+4 so this can be solved using the mean value theorem

Cause f(0)=4 and then you use that in the inequality etc.

[rustic heath]

But my problem is this.

Not how to actually prove the statement

[chilly mortar]

Here you definitely cannot reaso with <=>

you can however reason with simple implications

So if I reformulate your reasoning, the following holds

For any x, x < 2x

Therefore, since f' is decreasing, f'(x) > f'(2x)

Hence, 2 f'(x) - 2f'(2x) > 0 for ANY x

[rustic heath]

Yeah but why am i even allowed to use the derivative here?

[chilly mortar]

We just considered f' like a function like any other

that is decreasing

[rustic heath]

Ok so then when we get to the last part what do i do?

[chilly mortar]

You use integration

[rustic heath]

Oh im allowed to?

[chilly mortar]

since that is true for any x, it also holds for the following quantity

Hold on

Ah wait

Hmm

We need a point of evaluation first

like, a constant

[rustic heath]

There will be a constant introduced no?

Yeah

But also

Why can i use integration in this inequality

And not in an inequality like f'(x)>0<=>f(x)>0+c? Or can i?

[chilly mortar]

No

Basically, the idea is that if you take any nonnegative function g

the integral of g on any interval is also nonnegative

Take g(x) = 2 f'(x) - 2f'(2x)

[rustic heath]

Yeah

[chilly mortar]

So you can integrate it in an interval [a, t]

where a is fixed (we wil choose it) and t is a variable

and this function of t will be nonnegative

[rustic heath]

I don't see why tbh

[chilly mortar]

but we need to pick our a carefully

[chilly mortar]

well, if you integrate a nonnegative function, you compute a positive area

[rustic heath]

Oh we pick an a such that g will be nonnegative?

[rustic heath]

Ohh

Yeah

[chilly mortar]

So yeah

we just gotta pick our a decently

[rustic heath]

Is that true for indefinite integration too?

[chilly mortar]

No

That is why we will take an interval [a, t]

[rustic heath]

I see

Ok so

[chilly mortar]

So $\int_{a}^{t} (2f'(x) - 2f'(2x))dx > 0$

[rustic heath]

Returning to my original question, ig if i have an inequality f(x)>g(x) I cannot use the derivative?

[stable silo]

Rion

[rustic heath]

Yeah i see

That's pretty smart

[chilly mortar]

So now we just try to figure out what this is

It's nothing but

$2f(t) - 2f(a) - f(2t) + f(2a) > 0$

[stable silo]

Rion

[rustic heath]

Yeah we have f(0)=4

[chilly mortar]

Or in other words, $2f(t) - f(2t) > 2f(a) - f(2a)$

[stable silo]

Rion

[chilly mortar]

So we just gotta pick $a$ that suits our needs

[stable silo]

Rion

[chilly mortar]

(with a < t of course)

since here we care about positive t, I guess a = 0 can work

So, to conclude, by using a = 0 and knowing that f(0) = 4

it holds:

$2f(t) - f(2t) > f(0) = 4$

[stable silo]

Rion

[chilly mortar]

And this holds for ANY t > 0

[rustic heath]

Oh wait

I said x>0 hiw did we use f(0)?

[chilly mortar]

As said before, we integrated on an interval [a, t]

and decide what a and t are later

as long as a < t is fulfilled, it's fair game

So we apply that to a = 0, so the thing holds for any t > 0

[wicked wraith]

At the bottom what is next to the infinity sign i cannot tell lol

[rustic heath]

It's just (0,+∞) lol

[rustic heath]

Hmm

[wicked wraith]

Im such an idiot

[rustic heath]

Lmao

[wicked wraith]

thought that was a four lmfao

[rustic heath]

Lol

[chilly mortar]

Do you understand the reasoning?

[rustic heath]

Yeah but I don't understand why mine is necessarily wrong

Am i not allowed to use the derivative on inequalities?

[chilly mortar]

Because here you can definitely not go back the other way around

just because the integral of a function on an interval is positive

it does not mean that said function is positive on the interval

[rustic heath]

So in another case i could use it?

[chilly mortar]

Basically, g > 0 implies that the integral of g on an interval [a, b] is positive

which we used here

[rustic heath]

Yeah

[chilly mortar]

but the other way around is not true

so we can't get equivalence

you can only proceed using =>

[rustic heath]

I can?

[chilly mortar]

or rather, <=, if we follow the order of your thing

[rustic heath]

Oh right

Ok

[wicked wraith]

the inequality "2f(x) - f(2x) > 4" holds true for x > 0 so it is an inequality and f is strictly decreasing

[chilly mortar]

For mathematical reasoning, I generally do not recommend using => or <= symbools

It is better to write it as plain text, as I did before

[chilly mortar]

from here

[rustic heath]

Nor <=>?

[chilly mortar]

yeah, don't use that either

[rustic heath]

I literally only use this all the time without even thinking if it matters

[chilly mortar]

it will cost you very dearly on the long run

[rustic heath]

I mean in school we've never talked about this tbh

[chilly mortar]

I know, this is my advice as a senior

[rustic heath]

I'll try to get unused to it but i don't think i can lol

[chilly mortar]

back in high school my friends used <=> all the time

[rustic heath]

I'm actually writing exams to get to uni in a month so I'm wondering if using this could olay a part?

Could i lose points if i use it wrongly?

[chilly mortar]

yes, it will cost you the entire question in a question like this

[rustic heath]

What

[chilly mortar]

I will give you zero marks if you write that in an exam

for a question

It's not to be mean, I will explain to you why

[rustic heath]

Yeah ok i would too because it's incorrect tbf

[chilly mortar]

This problem is a lot more complex than it looks

And here, you can see that reasoning multi-step is required to solve this problem

and not in any order

So your ability to do so will determine how many points you will get

and clearly, using equivalence does NOT work in our case, it doesn't mean anything

therefore I cannot give you any point

[rustic heath]

Yeah tru

Got any other advice for exams?

[chilly mortar]

This is why, while I still can, I emphasize the importance of this

[chilly mortar]

For uni entrance exams?

[rustic heath]

Yeah

The format is

[chilly mortar]

Hmm let me think

[rustic heath]

3 hours with 4 exercises getting progressively harder (first is theory knowledge, like proofs definitions, true or false etc.)

The other three are exercises that involve stuff like this ig

[chilly mortar]

I remember that back in high school, our math speciality professor really underlined those tips for logical reasoning

as in, don't use arrows everywhere

[rustic heath]

Well same here

[chilly mortar]

But other than that, I would think of being as rigorous as you can

Make sure you verify the hypotheses to apply a theorem/result

make sure you name the theorem, if not trivial

[rustic heath]

Hmm

My main problem is time management i think

Like

I want to solve everything

[chilly mortar]

Very simple example: in a right triangle, a² + b² = c²

Here, name the theorem

[rustic heath]

Pythagorean

[chilly mortar]

Yeah

[wicked wraith]

2f(x) - f(2x) > 4

f'(x) > f'(2x)

2f(x) - f(2x) > 0

f(x) > 4

f(x) > C > 4

2f(x) - f(2x) > 4

its an inequality

[chilly mortar]

This is a non trivial result

[rustic heath]

Non trivial?

[chilly mortar]

Yes, the Pythagorean identity is not trivial, so you cannot pull it out of your ass

Or, if you do, name the theorem

[rustic heath]

I didn't think of this method i thought of mean value theorem tbh

[rustic heath]

Oh

Like what more do i have to do?

[chilly mortar]

As I said, if you use a theorem to derive a result, be sure to name it

in high school, for instance, the squeeze theorem can occur a lot

so be sure to name it if you use it, instead of just dropping the result

[rustic heath]

Not sure ik that one

[rustic heath]

Yeah i see what you're saying

[chilly mortar]

also known as the sandwich theorem

[rustic heath]

Lemme check if i know it

[chilly mortar]

No idea what that means

I see the top line and the bottom line are the same

so I don't see what exactly is being proven here

[rustic heath]

Ohh yeah ik it

[rustic heath]

Didn't he do the same thing?

[chilly mortar]

I don't know, the reasoning makes no sense

there's no order or logic whatsoever between the steps

[rustic heath]

Oh wait yeah isn't the integration wrong?

[chilly mortar]

the integration is correct but there is no justification as for why the integral is > 0

[rustic heath]

Wait wtf

[chilly mortar]

also it's missing the constants of integration

so yeah that's an example of what not to do

[rustic heath]

Lol

How do i become faster tho

Like istg that's my only problem

[chilly mortar]

Yeah, that's true

I guess you're also short on time

To be fair I cannot help much on that aspect

I am a very slow writer so I usually never finish my tests and exams

but that's my problem because my writing does not keep up with my thinking

[rustic heath]

Idk what my problem is

[chilly mortar]

I guess figuring that out could help you improve

[rustic heath]

I waste too much time in questions i can't necessarily solve i think

[chilly mortar]

if you need to be more familiar with your lessons, it's something that can still be helped with practice

but if you have a problem like me where your handwriting is too slow, idk how to help that

[rustic heath]

No it's pretty fast

Looks like crap tho

Anyway

[chilly mortar]

Hmm yeah that can be a problem

[rustic heath]

Thx for helping

[chilly mortar]

I would recommend, in general, do what you can

think about what you can't do after

[rustic heath]

There's no after😭

When i don't do something

[chilly mortar]

then it's fine, at least you'll be able to tell yourself, "I did what I could"

[rustic heath]

I never manage to see it again cause my time is up

[rustic heath]

Yeah but like

When i can't do something

Like my entire psychology falls apart

I can't bear knowing there was something i couldn't solve

Idk if it's an egoism thing

Or that I'm scared I'll do shit in the exam

[chilly mortar]

Just do your best and see what happens

It's not useful to think about what you can't do

If I were to compare mysef to Einstein at every of my setbacks I would not get anywhere

[rustic heath]

Lol tru tbf

[wicked wraith]

you have the inequality 2f(x) - f(2x) > 0 which means the difference between 2f(x) and f(2x) is more than zero, f(x) > 4 f(x), this means f(x) must be more than 4 for the first inequality to be correct, f(x) > C > 4, this adds a constant C where f(x) is more than C and C is more than 4, 2f(x) - f(2x) > 4, which proves that the difference between 2f(x) and f(2x) is obv more than 4, this proves its an inequality

[chilly mortar]

1. How did you get the initial inequality