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The Proof is Trivial!

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Reply 2420
Avatar for Flauta
Flauta
9
Original post by Khallil
All of the thanks goes to you for posting such great questions! (May I ask where you get them from?)

Also, have a look at LoTF's solution to problem 10. That substitution is another one of my favourites!

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The chapter on residue integration from advanced engineering mathematics. Was hoping to see someone use that in a solution, but these methods you guys keep posting are even sneakier :tongue:

Woah. Such a complicated question, would never have thought it could be reduced to something that simple. Will have to read through some of these solutions, might learn a thing or two from you guys' brilliance
Original post by henpen
Solution 390

Colour the plane black, then choose a point PP on none of the lines such that with each new line, switch the colour of all the regions that are on the same side of the new line as PP. If nn lines coincide, only perform this operation once.

I assume by adjacent you mean share an edge (rather than a vertex). If so, by passing across an edge you are passing over a line, but the colours on either side of the line must be different due to the operation we used to create it, therefore the adjacent regions have different colours.

This easily generalises to any number of dimensions.



I'm not sure if I am following you solution too well.

The question stated that lines are drawn on a plane

Then you said colour the plane and chose a point on none of those lines but at this point in your proof there are not any lines? Especially since the plane is black?

I didn't use the word adjacent? I spoke about regions but yes the would share an edge.

Then, no where did you mention the colours?
(edited 10 years ago)
Can you guys amend your problems/solutions' numeration (+1), starting from Henpen's differential equation?

Solution 391

We are gonna use induction. For 11 line, the problem is trivial; so, we assume the statement true for n−1n-1.
Now draw a line, to get nn lines. The only adjacent regions that does not satisfy the coloring property are exactly those divided by the new line. Choose one of the half-planes to which the nnth line divides the plane, and invert its coloring - in other words, if a region has been black, it is now red, and vice versa. Hence we are done.

Solution 392

It follows directly from the fact, the proof of which is of course inductive, that un=(n−1)!!u_{n}=(n-1)!!.
(edited 10 years ago)
Reply 2423
Avatar for henpen
henpen
Original post by Arieisit
I'm not sure if I am following you solution too well.

The question stated that lines are drawn on a plane

Then you said colour the plane and chose a point on none of those lines but at this point in your proof there are not any lines? Especially since the plane is black?

I didn't use the word adjacent? I spoke about regions but yes the would share an edge.

Then, no where did you mention the colours?


I'm aware, my solution was of the form of a method one could use to draw the lines on the plane such that no two adjacent regions have the same colour.

My point is that, given knowledge of where the lines lie, choose an arbitrary point not on any of them, remove all the lines, then redraw each line in an arbitrary order using my proposed method.

Indeed you didn't.

In colouring the plane black then inverting the colours of each half plane in turn.
Guys I know this question is probably very easy but I honestly have no clue what to do.
Question:
Let a,b,c be positive real numbers. Prove that

a^3 +b^3 +c^3 > ba^2 + cb^2 + ac^2

By the way its meant to be greater than or equal to. I tried using am-gm but kept missing that result. Thanks in advance.

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Reply 2425
Avatar for 0x2a
0x2a
2
Original post by theuser77
Guys I know this question is probably very easy but I honestly have no clue what to do.
Question:
Let a,b,c be positive real numbers. Prove that

a^3 +b^3 +c^3 > ba^2 + cb^2 + ac^2

By the way its meant to be greater than or equal to. I tried using am-gm but kept missing that result. Thanks in advance.

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By AM-GM
2a3+b3≥3a2b2a^3 + b^3 \geq 3a^2b

2b3+c3≥3b2c2b^3 + c^3 \geq 3b^2c

2c3+a3≥3c2a2c^3 + a^3 \geq 3c^2a


⇒3a3+3b3+3c3≥3a2b+3b2c+3c2a\Rightarrow 3a^3 + 3b^3 + 3c^3 \geq 3a^2b + 3b^2c + 3c^2a

∴a3+b3+c3≥a2b+b2c+c2a\therefore a^3 + b^3 + c^3 \geq a^2b + b^2c + c^2a .
(edited 10 years ago)
Original post by theuser77
Guys I know this question is probably very easy but I honestly have no clue what to do.
Question:
Let a,b,c be positive real numbers. Prove that

a^3 +b^3 +c^3 > ba^2 + cb^2 + ac^2

By the way its meant to be greater than or equal to. I tried using am-gm but kept missing that result. Thanks in advance.

Posted from TSR Mobile


Without loss of generality, a≤b≤ca \le b \le c. Hence a3+b3+c3≥a2b+b2c+c2aa^{3}+b^{3}+c^{3} \ge a^{2}b+b^{2}c+c^{2}a.
Sorry how did you get the first inequality

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Reply 2428
Avatar for 0x2a
0x2a
2
Original post by theuser77
Sorry how did you get the first inequality

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If you're talking about my post then

a3+a3+b33=2a3+b33≥a2b⇒2a3+b3≥3a2b\dfrac{a^3 + a^3 + b^3}{3} = \dfrac{2a^3 + b^3}{3} \geq a^2b \Rightarrow 2a^3 + b^3 \geq 3a^2b .
Original post by 0x2a
If you're talking about my post then

a3+a3+b33=2a3+b33≥a2b⇒2a3+b3≥3a2b\dfrac{a^3 + a^3 + b^3}{3} = \dfrac{2a^3 + b^3}{3} \geq a^2b \Rightarrow 2a^3 + b^3 \geq 3a^2b .



Sorry if I am asking really stupid question but how did you derive the first inequality from am-gm
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Reply 2430
Avatar for 0x2a
0x2a
2
Original post by theuser77
Sorry if I am asking really stupid question but how did you derive the first inequality from am-gm
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AM-GM with three variables states that

a+b+c3≥abc3\dfrac{a + b + c}{3} \geq \sqrt[3]{abc}.
But surely to get from
2a+b/3 >cube root (a^2b) to a^2b on the rhs you have to cube both sides
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Original post by theuser77
But surely to get from
2a+b/3 >cube root (a^2b) to a^2b on the rhs you have to cube both sides
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It is simply a3+a3+b3≥3a3a3b33=3a2ba^{3}+a^{3}+b^{3} \ge 3\sqrt[3]{a^{3}a^{3}b^{3}}=3a^{2}b.
Reply 2433
Avatar for 0x2a
0x2a
2
Original post by theuser77
But surely to get from
2a+b/3 >cube root (a^2b) to a^2b on the rhs you have to cube both sides
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Just replace the a and b with a^3 and b^3 and the result should be immediate.
Problem 393**

Prove that if a,b,c≥1a,b,c \ge 1, and a+b+c=9a+b+c=9, then ab+bc+ca≤a+b+c\sqrt{ab+bc+ca} \le \sqrt{a} + \sqrt{b}+\sqrt{c}.

Problem 394***

Find I=∫01ln⁡x⋅ln⁡2(1−x)xdx\displaystyle I = \int_{0}^{1} \frac{\ln x \cdot \ln^{2}(1-x)}{x} dx.

Problem 395***

Evaluate I=∫0∞(xex−e−x−12)⋅1x2dx\displaystyle I = \int_{0}^{\infty} \left(\frac{x}{e^{x}-e^{-x}} - \frac{1}{2} \right) \cdot \frac{1}{x^{2}}dx.

I have just noticed that there is no solution to problem 365. It is kinda ugly and painful in a sense that it requires too much case work, so I thought we can reduce it to:

Problem 365***(revisited)

Find ∫0∞xa(b+xc)ddx\displaystyle \int_{0}^{\infty} \frac{x^{a}}{(b+x^{c})^{d}}dx, when a≥0,b>0,c>0,d>a+1c\displaystyle a \ge 0, b>0, c>0, d > \frac{a+1}{c}. We also throw out the condition d∈Zd \in \mathbb{Z}.
Reply 2435
Avatar for Flauta
Flauta
9
Original post by Mladenov
.

Sorry if this is really stupid to ask, but by ln⁡2x\ln^2x, do you mean ln⁡(ln⁡x)\ln(\ln x) or (ln⁡x)2(\ln x)^2?
Original post by henpen
I'm aware, my solution was of the form of a method one could use to draw the lines on the plane such that no two adjacent regions have the same colour.

My point is that, given knowledge of where the lines lie, choose an arbitrary point not on any of them, remove all the lines, then redraw each line in an arbitrary order using my proposed method.

Indeed you didn't.

In colouring the plane black then inverting the colours of each half plane in turn.


I'm still having issues with your "proof" but here is my solution

Solution 390

Let P(n)P(n) be the statement of the result for nn lines.
Clearly, P(1)P(1) is true; colour one side of the line red then the other side black.
Assume P(n)P(n) is true.
Let a diagram be drawn with n+1n+1 lines.
Disregarding any one line, the regions may be coloured in the desired way, by P(n)P(n).
Now, add the last line.
Change the colour of all regions on the other side of the line, and leave the colours of all regions on the other side of the line.
If two regions share a boundary then they have opposite colours, whether their common boundry is part of the last line or not.

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(edited 10 years ago)
Original post by Flauta
Sorry if this is really stupid to ask, but by ln⁡2x\ln^2x, do you mean ln⁡(ln⁡x)\ln(\ln x) or (ln⁡x)2(\ln x)^2?


It means [ln⁡(1−x)]2[\ln(1-x)]^2
(edited 10 years ago)
Reply 2438
Avatar for Flauta
Flauta
9
Original post by Arieisit
It means [ln⁡(1−x)]2[\ln(1-x)]^2

Ooh okay, thanks!
Original post by Flauta
Sorry if this is really stupid to ask, but by ln⁡2x\ln^2x, do you mean ln⁡(ln⁡x)\ln(\ln x) or (ln⁡x)2(\ln x)^2?


As someone has said, it is (ln⁡(1−x))2(\ln(1-x))^{2}.

Original post by metaltron
Solution 393

By AM-GM

a+b+c≥9abc=(a+b+c)(abc)=a2bc+ab2c+abc2≥ab+bc+ca   (a,b,c≥1) \displaystyle \sqrt{a} + \sqrt{b} + \sqrt{c} \geq \sqrt{9abc} = \sqrt{(a+b+c)(abc)} = \sqrt{a^2bc + ab^2c + abc^2} \geq \sqrt{ab+bc+ca} \ \ \ (a,b,c \geq 1)


How do you derive your first inequality? It is not true. And, just by the way, this inequality is quite sharp.
(edited 10 years ago)

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