You should be able to solve this.

[Anonymous No.15539293]

Let [math]a_1, b_1, c_1[/math] be arbitrary positive numbers and for [math]n>1[/math] set

[math]a_{n+1} = b_n+1/c_n[/math]
[math]b_{n+1} = c_n+1/a_n[/math]
[math]c_{n+1} = a_n+1/b_n[/math]

Prove that at least one of [math]a_{800}, b_{800}, c_{800}[/math] is greater than 40 and that all three sequences are unbounded.

Include vax status with your answer.

[Anonymous No.15539320]

>>15539293
> all three sequences are unbounded
>Keep adding positive numbers
>wonder if it's unbounded
If you add infinitely many positive numbers, the series is unbound.
Vaxx status: pureblood

[Anonymous No.15539370]

>>15539293
Ummm sweaty letters can't be greater than numbers.
Vaxx status: triple booster Pfizer and Moderna

[Anonymous No.15540378]

wut

[Anonymous No.15540456]

>>15539320

That isn't true you fucktard, sum of 2^(-n) for all natural numbers n is equal to 2.

[Anonymous No.15540459]

>>15540456

Tripple Vaxxed

[Anonymous No.15540515]

>>15539293

I can "prove" the unboundedness thing, but it's a bit handwavy. First, ignore the fact that it's three chained sequences and instead consider the case of x_(n+1)=x_(n)+1/x_(n). One can clearly see that asymptotically, x_(n+x_(n)) ~= x_(n)+1, i.e. it takes n more applications of the equation to increase x_n by 1. So n is an exponential function of x_n, so x_n is a logarithmic function of n, so x_n is unbounded.

What happens if we consider the original problem—does this generalize? Well, consider the behavior of the sum of the variables (a_n+b_n+c_n). We find that the value of the sum after one iteration of the equations is itself plus the reciprocals of a_n, b_n, and c_n. All three of those reciprocals exceed the reciprocal of the sum itself, so we find that the sum must grow at least as quickly as x_n from the above function. As such it must also be unbounded. At least one of a_n, b_n, and c_n must equal or exceed a third of the sum, so at least one is also unbounded. WLOG, suppose it's a_n. a_n < c_(n+1), because b is always positive, so c is also unbounded. b is unbounded for the same reason. The end.

Triple-vaxxed.

[Anonymous No.15540556]

>>15539293

Okay, figured out the other part. So as before, note that the a_(n+1)+b_(n+1)+c_(n+1)=a_(n)+b_(n)+c_(n)+1/a_(n)+1/b_(n)+1/c_(n). Because 1/a_(n)+1/b_(n)+1/c_(n) is at least 3/(a_n+b_n+c_n) (as the harmonic mean is always lower than the regular mean), the sum is always at least the previous sum plus three over the previous sum.
Inverting this, we get that sum_(n)=(sum_(n+1)+-sqrt(sum_(n+1)^2-12))/2. This has no real solution for sum_n+1<sqrt(12). If we plug in sqrt(12) for sum_2 and programmatically calculate forward with our lower bound from like three sentences ago, we get that sum_800>=120.04263337534665>120, so at least one of the three variables exceeds (120/3=) 40 by n=800.

[Anonymous No.15540688]

>>15540556
>programmatically calculate forward
Cheating.

Now prove that [math]a_{789347832947242947298347} > 1256461565625[/math]

[Anonymous No.15540768]

>>15539320
1.111... < 1.2

[Anonymous No.15540844]

>>15539293
sqrt(2 * 800) is 40 as sqrt is concave down some is lost so the sum is greater than 40
3 * (40 + d) > 120
because f'x = (f x)^-1, sqrt(2x+2) is chosen to match the first term of sequence and sqrt(2*799+2) gives 40 as a lower bound. the first half of the proof just comes from giving the sum of a,b,c between iterations the benefit of the doubt and distributing the sum over the reciprocals, and before that assuming 40, 40, 40 containers. maybe.
vaxx status: vaxxed at least twice

[Anonymous No.15540886]

>>15540688
not them, person above, but, sqrt(2 * 789347832947242947298347+2) is greater than RHS

[Anonymous No.15540890]

>>15540886
*no +2, still true

[Anonymous No.15540898]

>there are TAs who have to read the incoherent, halfassed nonsense ITT and give it anything but a zero

[Anonymous No.15541602]

>>15539293
x=19, triangle is on the surface of a sphere.

[Anonymous No.15541655]

s(n) = a(n)+b(n)+c(n)
s(n+1) = a(n)+b(n)+c(n) + 1/a(n)+1/b(n)+1/c(n)

s(2) = (a(1)+1/a(1)) + (b(1)+1/b(1)) + (c(1)+1/c(1)) >= 2+2+2 = 6

by HM-AM 1/x+1/y+1/z >= 9/(x+y+z) so
s(n+1) >= s(n)+9/s(n)
s(n+1)^2/18 - s(n)^2/18 >= (s(n)+9/s(n))^2/18 - s(n)^2/18 > 1

s(N)^2/18 = (s(N)^2/18-s(N-1)^2/18) + ... + (s(3)^2/18-s(2)^2/18)+s(2)^2/18 > N-2 + 6^2/18 = N

s(N) > 3 sqrt(2N)

[Anonymous No.15541743]

>>15539293
>>15540556
You can do better.
1/a + 1/b + 1/c >= 9/(a+b+c). This can easily be shown using convexity of 1/x.
Sum(n+1) >= Sum(n) + 9/Sum(n)
You can use means instead of sums to get:
Mu(n+1) >= Mu(n) + 1/Mu(n)
Squaring both sides gives:
Mu(n+1)^2 >= Mu(n)^2 + 2 + 1/Mu(n)^2
Assuming Mu(n)>=1 for n>1, Mu(n+1)^2 > Mu(n)^2 + 2.
This gives Mu(n+k)^2 > Mu(n)^2 + 2k for n>1
For n=1, Mu(2)^2 >= Mu(1)^2 + 2 + 1/Mu(1)^2 >= 2+2 since x+1/x has minimum 2
Altogether, Mu(n)^2 > Mu(2)^2 + 2*(n-2) >= 2n for n>2
Mu(n) > sqrt(2n)
Sum(n) > 3*sqrt(2n)

Plugging in n=800 gives Sum(800) > 3*40 which means at least 1 term > 40.
In general Sum(n) will have at least 1 term > sqrt(2n). This term will cycle through a,b,c at n,n+1,n+2 with a little added each iteration which means all 3 sequences are unbounded.

Pureblood btw

[Anonymous No.15541870]

Assume all three sequences a_n, b_n, c_n converge to a, b, and c respectively.

This lets us write the equations without indices in their n-> inf limit, so
a = b + 1/c
a = c + 1/a + 1/c
c = c + 1/a + 1/b + 1/c
=> 0 = 1/a + 1/b + 1/c

This is a contradiction, so they cannot all converge. But we know if any of the diverge, then they all diverge, so they must all diverge.

fully vaxxed

[Anonymous No.15542377]

>>15539293
its the square root of 19

[Anonymous No.15542756]

>>15539293
fully vaxed, plugged into excel and numerically calculated with A = B = C = 1, it works. am engineer, dont care for proofs