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The Proof is Trivial!

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Reply 560
Avatar for Jkn
Jkn
11
Original post by Mladenov
You are right, the points, at which the partial derivatives fail to exist, are good candidates for extremums. Yet, this does not help us at all.
I tried several methods to prove your inequalities, including some brute-force techniques such as Lagrange Multipliers, but nothing works. Also, when a=1,b=7,c=8a=1,b=7,c=8, we have f(1,7,8)<32=f(1,1,1)f(1,7,8)< \frac{3}{2} = f(1,1,1). Hence f(a,b,c)≥32af(a,b,c) \ge \frac{3}{2a} for all a,b,c>0a,b,c > 0 is clearly not true. I have not tried f(a,b,c)≥32cf(a,b,c) \ge \frac{3}{2c}, but even if it is true, then you have to prove that RHS≤32cRHS \le \frac{3}{2c} is order your solution to work.

The problem is that you are doing the following:
Say you have correctly proven that for each x∈Sx \in S, f(x)≥g(x)f(x) \ge g(x). Then, you are claiming that f(x)≥max⁡(g(x)x∈S)f(x) \ge \max(g(x)_{x\in S}), which is generally not true.

Spoiler


Ah hmm that's what I feared, thanks for pointing that out.
Original post by Mladenov
Precisely. There is a solution using exactly these two inequalities.

I suspected this was the solution you would have been after but found it difficult to arrange the terms in useful sets. Non the less, I shall persist!
Reply 561
Avatar for Mladenov
Mladenov
1
Solution 73

I assume n≥3n \ge 3. Let SS be a finite multi-set of nn positive integers. If all elements of SS are greater than 11, we know that the product of all the elements of SS is greater than or equal to their sum.
Suppose that we have solution with exactly kk components greater than 11. Thus the remaining n−kn-k components are equal to 11. Now, the sum of the elements, which are greater than 11, is less than or equal to the product; the sum of the n−kn-k 11's is less than or equal to n−1n-1. It follows that k=1k=1.
As we are counting ordered sets, we have n(m−1)+1n(m-1)+1 solutions in positive integers.
The number of non-negative integer solutions of the equation x1+..+xn−1=n−1x_{1}+..+x_{n-1} = n-1 is (2n−3n−1)\dbinom{2n-3}{n-1}. Hence, the number of solutions in non-negative integers is 2(2n−3n−1)+n(m−1)+12\dbinom{2n-3}{n-1}+n(m-1)+1.

Problem 74 hint:

Spoiler

(edited 11 years ago)
Reply 562
Avatar for Jkn
Jkn
11
Original post by Mladenov
Solution 73

I assume n≥3n \ge 3. Let SS be a finite multi-set of nn positive integers. If all elements of SS are greater than 11, we know that the product of all the elements of SS is greater than or equal than their sum.
Suppose that we have solution with exactly kk components greater than 11. Thus the remaining n−kn-k components are equal to 11. Now, the sum of the elements, which are greater than 11, is less than or equal to the product; the sum of the n−kn-k 11's is less than or equal to n−1n-1. It follows that k=1k=1.
As we are counting ordered sets, we have n(m−1)+1n(m-1)+1 solutions in positive integers.
The number of non-negative integer solutions of the equation x1+..+xn−1=n−1x_{1}+..+x_{n-1} = n-1 is (2n−3n−1)\dbinom{2n-3}{n-1}. Hence, the number of solutions in non-negative integers is 2(2n−3n−1)+n(m−1)+12\dbinom{2n-3}{n-1}+n(m-1)+1.

Just when I thought no-one was interested. Exactly what I had in mind, nicely done! :smile:
Btw, I'm making some progress with your problem now using

Spoiler

which I wasn't familiar with until now (hopefully just a few algebraic tricks away from solution!)
Solution 61

Assume ff continuous (perhaps this was not necessary).

Let P=(0,1),  Q=(1,∞)\mathcal{P}=(0,1),\;\mathcal{Q}=(1,\infty), noting that f(1)=1f(1)=1. If f(s)=1f(s)=1 for some s∈Ps\in\mathcal{P}, then for all y,  f(sy)=1⇒f(r)=1y,\;f(s^y)=1 \Rightarrow f(r)=1 over P\mathcal{P}. However, then for y∈P:  f(xy)=f(x)⇒f=Cy\in \mathcal{P}:\;f(x^y)=f(x) \Rightarrow f=\mathcal{C}, and hence f=1f=1 over R+\mathbb{R}^+. Similarly if f(s)=1f(s)=1 for some s∈Qs\in\mathcal{Q} then f=1f=1 over R+\mathbb{R}^+. In a similar vein, suppose there exist distinct x0,y0∈Px_0,y_0\in \mathcal{P} such that f(x0)=f(y0)≠1f(x_0)=f(y_0)\neq 1 then f(x0)f(r)=f(y0)f(r)⇒f(x0r)=f(y0r)f(x_0)^{f(r)}=f(y_0)^{f(r)} \Rightarrow f(x_0^r)=f(y_0^r) which implies f(x)=Cf(x)=\mathcal{C} and since C=CC⇒C=±1\mathcal{C}=\mathcal{C}^{ \mathcal{C}}\Rightarrow \mathcal{C}=\pm 1 we have f=1f=1 over P\mathcal{P}, which, as previously demonstrated, implies f=1f=1 over R+\mathbb{R}^+. Similarly for the case x0,x1∈Qx_0,x_1\in\mathcal{Q}. Hence if ff is not injective it is 11 everywhere.

Naturally, we now assume ff is injective. Observe that f(zyx)=f(z)f(y)f(x)⇒f(zyn)=f(z)f(y)f(n)f(z^{y^x})=f( z)^{f(y)^{f(x)}}\Rightarrow f(z^{y^n})=f( z)^{f(y)^{f(n)}} but also that inductively, we have f(zyn)=f(z)f(y)nf(z^{y^n})=f(z)^{f(y)^n}. Combining these facts, f(n)=nf(n)=n for positive integers nn. Hence we have f(xn)=f(x)nf(x^n)=f(x)^n. Now suppose that f(x)>xf(x)>x over (a,b)⊂Q,(a,b)\subset\mathcal{Q}, then by the preceding statement, f(xn)>xnf(x^n)>x^n over (an,bn)(a^n,b^n) and hence the inequality is satisfied over arbitrarily large intervals, which contradicts f(n)=nf(n)=n. Similarly if f(x)<xf(x)<x over any subinterval of Q\mathcal{Q}. Hence we must have f(x)=xf(x)=x over Q\mathcal{Q}. Now assuming f(x)>xf(x)>x over P\mathcal{P}, our preceding proposition would imply that for y∈Qy\in\mathcal{Q} and x∈P:  yx=f(yx)<f(y)f(x)=f(yx)x\in\mathcal{P}:\; y^x=f(y^x)<f(y)^{f(x)}=f(y^x) which is absurd. Similarly if f(x)<xf(x)<x over P\mathcal{P}. Hence f(x)=xf(x)=x over P\mathcal{P} as well.

Therefore f(x)=1f(x)=1 or f(x)=x.f(x)=x.
Reply 564
Avatar for Mladenov
Mladenov
1
Original post by Lord of the Flies
...


Splendid solution. :tongue:
You are not using continuity, so it is not necessary.

Another approach:

Spoiler

Reply 565
Avatar for Jkn
Jkn
11
(Still can't do problem 74 (fml!) so thought "if you can't beat them, join them")

Problem 75

Show that 1b(a+b)+1c(b+c)+1a(c+a)≥32 \frac{1}{b(a+b)} + \frac{1}{c(b+c)} + \frac{1}{a(c+a)} \geq \frac{3}{2} for positive real numbers a,b,c such that abc=1 abc=1 .
Reply 566
Avatar for Mladenov
Mladenov
1
Solution 75

Let a=xy\displaystyle a=\frac{x}{y}, b=yz\displaystyle b=\frac{y}{z}, c=zx\displaystyle c=\frac{z}{x}.

We have 1b(a+b)+1c(b+c)+1a(c+a)=∑cycx4x3y+z2x2≥(x2+y2+z2)2∑cycx3y+∑cycx2y2\displaystyle \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)} = \sum_{cyc} \frac{x^{4}}{x^{3}y+z^{2}x^{2}} \ge \frac{(x^{2}+y^{2}+z^{2})^{2}}{ \sum_{cyc} x^{3}y + \sum_{cyc} x^{2}y^{2}}.
Since 2(x2+y2+z2)2=2∑cycx4+4∑cycx2y2\displaystyle 2(x^{2}+y^{2}+z^{2})^{2} = 2\sum_{cyc} x^{4}+ 4\sum_{cyc} x^{2}y^{2}, from
∑cycx4≥∑cycx3y\displaystyle \sum_{cyc} x^{4} \ge \sum_{cyc} x^{3}y (AM-GM), and
∑cyc(x4+x2y2)≥∑cyc2x3y\displaystyle \sum_{cyc} (x^{4}+x^{2}y^{2}) \ge \sum_{cyc} 2x^{3}y (obvious), we obtain
(x2+y2+z2)2∑cycx3y+∑cycx2y2≥32\displaystyle \frac{(x^{2}+y^{2}+z^{2})^{2}}{ \sum_{cyc} x^{3}y + \sum_{cyc} x^{2}y^{2}} \ge \frac{3}{2}

Spoiler

(edited 10 years ago)
Original post by joostan
This may be a silly question but does anybody like geometry?? Even Mladenov who appears to be a maths genius dislikes it :s-smilie:

Geometry is wonderful, if you can do it :tongue:
Geometry is also taught extremely badly in most countries.
Reply 568
Avatar for Jkn
Jkn
11
Original post by Mladenov

∑cycx4≥∑cycx3y\displaystyle \sum_{cyc} x^{4} \ge \sum_{cyc} x^{3}y (AM-GM), and
∑cyc(x4+x2y2)≥∑cyc2x3y\displaystyle \sum_{cyc} (x^{4}+x^{2}y^{2}) \ge \sum_{cyc} 2x^{3}y (obvious)

I was wondering if you'd use homogenisation! :smile:
But could you please explain how you have used AM-GM in the first of the two inequalities?
And why is the second inequality obvious? It rests on ∑cyc(x2y2)≥∑cycx3y\displaystyle \sum_{cyc} (x^{2}y^{2}) \ge \sum_{cyc} x^{3}y and combination with the first result.

(sorry If this is actually obvious to you, but I don't find it to be so!)

Spoiler

(edited 10 years ago)
Original post by Llewellyn
Geometry is wonderful, if you can do it :tongue:
Geometry is also taught extremely badly in most countries.


In Chinese high schools you are taught geometry endlessly, in exchange you don't learn things like integration (even basic integrals). That might perhaps be because China has a strong tradition of gearing school pupils towards the IMO and similar national-level contests, for which learning geometry instead of integration is perfectly justified.
Reply 570
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Jkn
11
Such a shame this thread is becoming a bit stagnant! All we need is 24 more problems before we can whip out the jay-z jokes!

A problem people may recognise:

Problem 76*/**

For positive real numbers a,b,c, prove that

(a2+b2)2≥(a+b+c)(a+b−c)(b+c−a)(c+a−b) \displaystyle(a^2+b^2)^2 \geq (a+b+c)(a+b-c)(b+c-a)(c+a-b)
(edited 10 years ago)
Reply 571
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Jkn
11
And another...

Problem 77*

Long John Silverman has captured a treasure map from AdamMcBones.
Adam has buried the treasure at the point (x,y) with integer co-ordinates (not necessarily positive).
He has indicated on themap the values of x2+yx^2 + y and x+y2x+y^2, and these numbers are distinct.

Prove that Long John has to dig only in one place to find the treasure.
(edited 10 years ago)
Original post by Mladenov
Splendid solution. :tongue:
You are not using continuity, so it is not necessary.

Another approach:

Spoiler



Ha! I did not think of extracting Cauchy from it, très bien! I mentioned continuity as a safety net in case I had overlooked something at that time of night :biggrin:


I should really be working, mais bon.

Solution 52

Note that since (an,an−1)(an−1,an−2)an+1=an−1[1+(an−1,an−2)]+an−2(a_n,a_{n-1})(a_{n-1},a_{n-2})a_{n+1}=a_{n-1}\big[1+(a_{n-1},a_{n-2})\big]+a_{n-2} we have (an,an−1)∣an−2(a_{n},a_{n-1})|a_{n-2} which then implies (an,an−1)∣(an−1,an−2)(a_{n},a_{n-1})|(a_{n-1},a_{n-2}) and inductively (an,an−1)∣(am,am−1)(a_n,a_{n-1})|(a_m,a_{m-1}) for any m≤nm\leq n. Hence there must be some n0n_0 such that n≥n0⇒(an,an−1)=Dn\geq n_0\Rightarrow (a_{n},a_{n-1})=\mathcal{D}. Then clearly for all n≥n0n\geq n_0 we must have an=Dqna_n=\mathcal{D} q_n where (qi,qj)=1(q_i,q_j)=1 for any i,j≥n0i,j\geq n_0. But since (an)(a_n) is bounded (qn)(q_n) is bounded as well, and hence we require qn≥N=1q_{n\geq N}=1 for some N≥n0N\geq n_0. Thus an≥N=Da_{n\geq N}=\mathcal{D}, which, if plugged into the definition of (an)(a_n) immediately gives an≥N=2a_{n\geq N}=2. Now we work backwards to show that all preceding terms are also 22. Obviously (aN,aN−1)∈{1,2}(a_N,a_{N-1})\in \{1,2\}, and given that 2(aN,aN−1)=2+aN−12(a_N,a_{N-1})=2+a_{N-1} we clearly cannot have (aN,aN−1)=1(a_N,a_{N-1})=1 and hence it follows that aN−1=2a_{N-1}=2 etc. down to a1=2a_1=2.

Therefore the only possible sequence is an=2,  ∀na_n=2,\;\forall n.



Original post by Jkn

But could you please explain how you have used AM-GM in the first of the two inequalities?
And why is the second inequality obvious?


The first: rearrangement ineq. for (wlog) x≥y≥zx\geq y\geq z and x3≥y3≥z3x^3\geq y^3\geq z^3. The second is just ∑cycx2(x−y)2≥0\displaystyle \sum_{\text{cyc}}x^2(x-y)^2\geq 0
By the way, in Problem 75, notice that by plugging in a couple coefficients into the inequality and taking the exact same approach as in Mladenov's solution we can easily generalise the result:

∑a,b,c1b(aλ+bμ)≥(x2+y2+z2)2(λ∑cycx3y+μ∑cycx2y2)−1\displaystyle \sum_{a,b,c}\frac{1}{b(a\lambda+b\mu)} \geq\Big( x^2+y^2+z^2\Big)^2 \left(\lambda \sum_{\text{cyc}}x^3y+\mu \sum_{\text{cyc}}x^2y^2\right)^{-1}

Then using the same tricks to show that λ(x2+y2+z2)2≥3λ∑cycx3y\displaystyle \lambda(x^2+y^2+z^2)^2\geq 3\lambda\sum_{\text{cyc}}x^3y and μ(x2+y2+z2)2≥3μ∑cycx2y2\displaystyle \mu(x^2+y^2+z^2)^2\geq 3\mu\sum_{\text{cyc}}x^2y^2 we get:
1b(aλ+bμ)+1c(bλ+cμ)+1a(cλ+aμ)≥3λ+μ\displaystyle \dfrac{1}{b(a\lambda+b\mu)}+ \dfrac{1}{c(b\lambda+c\mu)}+ \dfrac{1}{a(c\lambda+a\mu)}\geq \dfrac{3}{\lambda+\mu}



For those how dislike inequalities with a passion:

Problem 78 (a classic!)

Evaluate 1+21+31+4⋯\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\cdots}}}}

Problem 79

This is lolz-maths, but whatever. Let's see who can do it the fastest!

If f(x)=6x7ex2sin⁡2(x1000)f(x)=6x^7e^{x^2}\sin^2 (x^{1000}) find f(2013)(0)−f(2012)(0)f^{(2013)}(0)-f^{(2012)}(0)
(edited 10 years ago)
Reply 573
Avatar for james22
james22
16
Original post by Jkn
And another...

Problem 77*

Long John Silverman has captured a treasure map from AdamMcBones.
Adam has buried the treasure at the point (x,y) with integer co-ordinates (not necessarily positive).
He has indicated on themap the values of x2+yx^2 + y and x+y2x+y^2, and these numbers are distinct.

Prove that Long John has to dig only in one place to find the treasure.


I've done this before so will give someone else a try before I post but that is a very nice question.
Reply 574
Avatar for james22
james22
16
Original post by Lord of the Flies

Problem 79

This is lolz-maths, but whatever. Let's see who can do it the fastest!

If f(x)=6x7ex2sin⁡2(x1000)f(x)=6x^7e^{x^2}\sin^2 (x^{1000}) find f(2013)(0)−f(2012)(0)f^{(2013)}(0)-f^{(2012)}(0)


Does f^(n) mean the nth derivative or f applied n times. I think its the former but if its the latter than answer is 0.
Original post by james22
Does f^(n) mean the nth derivative or f applied n times. I think its the former but if its the latter than answer is 0.


Yes, nth derivative.
Reply 576
Avatar for Mladenov
Mladenov
1
Original post by Jkn
..


AM-GM - ∑cycx4≥∑cycx2y2\displaystyle \sum_{cyc} x^{4} \ge \sum_{cyc} x^{2}y^{2} and ∑cyc(x4+x2y2)≥2∑cycx3y\displaystyle \sum_{cyc} (x^{4}+x^{2}y^{2}) \ge 2\sum_{cyc} x^{3}y, imply ∑cycx4≥∑cycx3y\displaystyle \sum_{cyc} x^{4} \ge \sum_{cyc} x^{3}y.
Problem 74 hint

Spoiler



Original post by Lord of the Flies
Solution 52...


C'est ça!

Solution 76

We can suppose a+b−c>0a+b-c > 0, a+c−b>0a+c-b >0, and b+c−a>0b+c-a >0. AM-GM and C-S yield (a+b−c)2(a+c−b)(b+c−a)≤(2a+2b)444<(a2+b2)2(a+b-c)^{2}(a+c-b)(b+c-a) \le \frac{(2a+2b)^{4}}{4^{4}} < (a^{2}+b^{2})^{2}.
There are two possibilities: either my solution is totally wrong, or this is too weak inequality.

Solution 77

Suppose for the sake of contradiction that there are four distinct integers a,b,c,da,b,c,d such that
a2+b=c2+da^{2}+b=c^{2}+d;
a+b2=c+d2a+b^{2}=c+d^{2}.
(a−c)(a+c)=d−b(a-c)(a+c)=d-b, a−c=d2−b2a-c=d^{2}-b^{2}. Thus, a+c=d+b=±1a+c=d+b= \pm1. The case −1-1 implies c=−1−ac=-1-a, d=−1−bd=-1-b. So, a2+b=c2+d=1+2a+a2−1−ba^{2}+b=c^{2}+d= 1+2a+a^{2}-1-b, a=ba=b-contradiction. Let a+c=b+d=1a+c=b+d=1. We then have a−c=d−ba-c=d-b and therefore a=da=d, b=cb=c. Consequently, a2+b=d2+c=a+b2a^{2}+b=d^{2}+c=a+b^{2}-contradiction. Hence our person (I do not remember his name :biggrin:) has to dig in only one place.

Solution 78

Define f(x)=1+xf(x+1)f(x)=\sqrt{1+xf(x+1)}. It is quite well-known that f(x)=x+1f(x)=x+1. Hence 1+21+31+4⋯=3\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt {\cdots}}}}=3.
There are several ways to solve this functional equation. We can bound ff above and below; or we can introduce the function g(m,n)=1+(m−1)g(m+1,n)g(m,n) = \sqrt{1+(m-1)g(m+1,n)} and our expression is lim⁡n→∞g(3,n)\displaystyle \lim_{n \to \infty} g(3,n); finally, we can consider the general equation f(x)=(g(x)+xαf(x+a))1β\displaystyle f(x) = (g(x)+x^{\alpha}f(x+a))^{\frac{1}{\beta}}, where x,a,α≥0x,a, \alpha \ge 0, β>1\beta >1, and g:R+∪0→R+∪0g : \mathbb{R^{+}} \cup 0 \to \mathbb{R^{+}} \cup 0, inf⁡g>0\inf g > 0, and solve it iteratively.
Reply 577
Avatar for Jkn
Jkn
11
Original post by Lord of the Flies

Problem 75, notice that by plugging in a couple coefficients into the inequality and taking the exact same approach as in Mladenov's solution we can easily generalise the result:

∑a,b,c1b(aλ+bμ)≥(x2+y2+z2)2(λ∑cycx3y+μ∑cycx2y2)−1\displaystyle \sum_{a,b,c}\frac{1}{b(a\lambda+b\mu)} \geq\Big( x^2+y^2+z^2\Big)^2 \left(\lambda \sum_{\text{cyc}}x^3y+\mu \sum_{\text{cyc}}x^2y^2\right)^{-1}

Then using the same tricks to show that λ(x2+y2+z2)2≥3λ∑cycx3y\displaystyle \lambda(x^2+y^2+z^2)^2\geq 3\lambda\sum_{\text{cyc}}x^3y and μ(x2+y2+z2)2≥3μ∑cycx2y2\displaystyle \mu(x^2+y^2+z^2)^2\geq 3\mu\sum_{\text{cyc}}x^2y^2 we get:
1b(aλ+bμ)+1c(bλ+cμ)+1a(cλ+aμ)≥3λ+μ\displaystyle \dfrac{1}{b(a\lambda+b\mu)}+ \dfrac{1}{c(b\lambda+c\mu)}+ \dfrac{1}{a(c\lambda+a\mu)}\geq \dfrac{3}{\lambda+\mu}



I could see how it could be come my rearrangement, just have no idea how it could've come from AM-GM (Directly at least)! Ah it really is obvious now that you put it like that, cheers!

Holy crap, what an insight! The result looks so much more beautiful that way :smile:

Problem 78 (a classic!)

Evaluate 1+21+31+4⋯\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\cdots}}}}

Love this question! :biggrin: Kind of makes me wish I was born 100 years ago so I could've send a solution off to the magazine! (Ramanujan had to publish the solution himself I think. Deriving it from the more generalised version)
(edited 10 years ago)
Reply 578
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Jkn
11
Original post by Mladenov
AM-GM - ∑cycx4≥∑cycx2y2\displaystyle \sum_{cyc} x^{4} \ge \sum_{cyc} x^{2}y^{2} and ∑cyc(x4+x2y2)≥2∑cycx3y\displaystyle \sum_{cyc} (x^{4}+x^{2}y^{2}) \ge 2\sum_{cyc} x^{3}y, imply ∑cycx4≥∑cycx3y\displaystyle \sum_{cyc} x^{4} \ge \sum_{cyc} x^{3}y.

But your solution says ∑cycx4≥∑cycx3y\displaystyle \sum_{cyc} x^{4} \ge \sum_{cyc} x^{3}y by AM-GM[QUOTE]. Can I assume you meant x^2y^2?

Problem 74 hint

Spoiler



ahhhhhh!

C'est ça!

Solution 76

We can suppose a+b−c>0a+b-c > 0, a+c−b>0a+c-b >0, and b+c−a>0b+c-a >0. AM-GM and C-S yield (a+b−c)2(a+c−b)(b+c−a)≤(2a+2b)444<(a2+b2)2(a+b-c)^{2}(a+c-b)(b+c-a) \le \frac{(2a+2b)^{4}}{4^{4}} < (a^{2}+b^{2})^{2}.
There are two possibilities: either my solution is totally wrong, or this is too weak inequality.

I'm not quite sure how you got rid of the c using Am-Gm! In any case observing the behaviour of the inequality as c approaches infinity is sufficient is disproving the inequality. Your mistake, of course, being that a+b−c>0 a+b-c>0 is not valid! I'm sure you simply skimmed over it thinking you need to apply AM-GM and C-S, but I think this question is more about dealing with the "c" and

Spoiler

than anything else (though you may find a different solution to me!) :smile:
Solution 78

Define f(x)=1+xf(x+1)f(x)=\sqrt{1+xf(x+1)}. It is quite well-known that f(x)=x+1f(x)=x+1. Hence 1+21+31+4⋯=3\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt {\cdots}}}}=3.
There are several ways to solve this functional equation. We can bound ff above and below; or we can introduce the function g(m,n)=1+(m−1)g(m+1,n)g(m,n) = \sqrt{1+(m-1)g(m+1,n)} and our expression is lim⁡n→∞g(3,n)\displaystyle \lim_{n \to \infty} g(3,n); finally, we can consider the general equation f(x)=(g(x)+xαf(x+a))1β\displaystyle f(x) = (g(x)+x^{\alpha}f(x+a))^{\frac{1}{\beta}}, where x,a,α≥0x,a, \alpha \ge 0, β>1\beta >1, and g:R+∪0→R+∪0g : \mathbb{R^{+}} \cup 0 \to \mathbb{R^{+}} \cup 0, inf⁡g>0\inf g > 0, and solve it iteratively.

Dayyyyyum son you post fast!
Reply 579
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Mladenov
1
Original post by Jkn
But your solution says ∑cycx4≥∑cycx3y\displaystyle \sum_{cyc} x^{4} \ge \sum_{cyc} x^{3}y by AM-GM. Can I assume you meant x^2y^2?


Nope, I meant x3yx^{3}y. From AM-GM we have ∑cycx4≥∑cycx2y2\displaystyle \sum_{cyc} x^{4} \ge \sum_{cyc} x^{2}y^{2} and ∑cyc(x4+x2y2)≥2∑cycx3y\displaystyle \sum_{cyc} (x^{4}+x^{2}y^{2}) \ge 2\sum_{cyc} x^{3}y, the second is also trivially true, and hence ∑cycx4≥∑cycx3y\displaystyle \sum_{cyc} x^{4} \ge \sum_{cyc} x^{3}y. It is a consequence of AM-GM, yet not a direct result.

Problem 76 is wrong in this form (I thought c≤a+bc \le a+b).

Edit: We have to bound cc; I can try to find the best possible cc.

Solution 79

As x→0x\to 0, from Taylor's series we have sin⁡(x)=x+O(x3)\sin (x) = x+\mathcal{O}(x^{3}), ex=1+x+12x+16x3+O(x4)e^{x}=1+x+\frac{1}{2}x+\frac{1}{6}x^{3}+ \mathcal{O}(x^{4}).
Thus, 6x7ex2sin⁡(x1000)=6x7(x1000+O(x2000))2(1+x2+12x4+16x6+O(x8))6x^{7}e^{x^{2}}\sin (x^{1000})= 6x^{7}(x^{1000}+\mathcal{O}(x^{2000}))^{2}(1+x^{2}+\frac{1}{2}x^{4}+\frac{1}{6}x^{6}+\mathcal{O}(x^{8})).
Hence f(2013)(0)−f(2012)(0)=2013!f^{(2013)}(0)-f^{(2012)}(0)= 2013!


Problem 80

Let (an)n≥1(a_{n})_{n\ge1} be an increasing sequence of positive integers. Suppose also that lim⁡n→∞ann=0\displaystyle \lim_{n\to \infty} \frac{a_{n}}{n}=0. Then, the sequence bn=nan\displaystyle b_{n} = \frac{n}{a_{n}} contains all positive integers.
(edited 10 years ago)

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