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be positive reals. Prove that there is a cyclic permutation 
of 
such that the inequality:![\[
\frac{a}{xa + yb + zc} + \frac{b}{xb + yc + za} + \frac{c}{xc + ya + zb} \geq \frac{3}{x + y + z}
\]](http://latex.artofproblemsolving.com/5/5/3/553527fc40a1b7d29dbdfb4b5e63635308f46c85.png)
and 
.
denote the set of all real numbers. Find all functions 
such that
for all real numbers 
, 
.
+1 w
such that every collection of consecutive positive integers contains an integer whose sum of digits is divisible by 
. for which he can do this, or prove that none exist.
be two positive sequences defined by 
and![\[ \begin{cases} {{x}_{n+1}}{{y}_{n+1}}-{{x}_{n}}=0 \\ x_{n+1}^{2}+{{y}_{n}}=2 \end{cases} \]](http://latex.artofproblemsolving.com/2/0/1/2012dc6ff3a41478547cea4aa9b6bccbe17a3623.png)
for all 
.
and 
for some 
. Then 
, so 
(the sequences have positive terms), and then 
.
and 
, the conclusion painlessly follows (the limits are 
, respectively 
).
and 
, meaning we can take 
and 
, by induction we will get 
and 
.
and 
. Let 
-- observe 
. We claim that 
Indeed, verify by double angle that 
and 
by the double angle formula. But 
and 
, so as 
approaches 0, the two limits are 0 and 2.
and 
for all 
, so 
, 
, and we have 
and 
. Recognizing this, as well as the fact that 
and 
, we now claim that 
and 
. This holds, as we have 
and 
by double angle. Then 
, and likewise 
. 
.
for all positive integers 
. (base case 
is obvious). Suppose it was true for everything up to 
. We show 
.
, and ![\[y_{t+1}^2 = \frac{x_t^2}{x_{t+1}^2} = \frac{4 - y_t^2}{2 - y_t} = 2 + y_t\]](http://latex.artofproblemsolving.com/3/d/0/3d0cc1360be9cd21f21a457143128e806abac682.png)
Hence 
. 
, where 
for all positive integers .
, so 
. Since the sequence is positive, we have 
by half angle.
, we have 
, so 
tends to and 
tends to .
, 
, 
, and 
. So we want to prove that 
, 
., are as above, and let's show that this pattern holds for 
, 
.![\[x_{n+1}^2=2-y_n=2-2\cos\left(\frac{30}{2^{n-1}}\right)\Longleftrightarrow x_{n+1}=2\sin\left(\frac{30}{2^n}\right).\]](http://latex.artofproblemsolving.com/7/8/7/787f52233ab7449978ea21bdad1a14643af8a42b.png)
Also use the first equation to get 
so![\[\lim_{n\rightarrow\infty}x_n=0 \text{ and }\lim_{n\rightarrow \infty}y_n=2.\]](http://latex.artofproblemsolving.com/f/b/b/fbb6232edf9a9fee26e908e1a32783e1c52b5c06.png)
and 
. We will prove this by induction on . For 
, this is true by definition. Else, assume it holds for ; we will show it holds for 
. Using the second given equation, we have
By the sine half angle formula, this implies that 
For 
, we use the first given equation to get
using the sine double angle formula. This completes the induction.
and 
.
and 
, as well as 
and 
. Now, we proceed by induction. Suppose that 
and 
, then ![\[x^2_{n+1}=2-y_n = 2-2\cos(90-\theta) = 2-2(2\cos^2(45-\theta/2)-1)=4\sin^2(45-\theta/2),\]](http://latex.artofproblemsolving.com/9/3/b/93b2573293f163975b677ce29d7f0ceb3274fbab.png)
giving 
. Now, since 
, we see that 
works because ![\[2\sin(45+\theta/2)\cos(45+\theta/2) = \sin(90+\theta)=\cos(-\theta)=\cos(\theta).\]](http://latex.artofproblemsolving.com/9/0/b/90b9c8cbb15aff23bf111dcb2464b74f015ddcf0.png)
Thus, ![\[(x_n,y_n)=(2\cos\theta, 2\sin\theta)\implies (x_{n+1},y_{n+1}) = (2\cos(45+\theta/2), 2\sin(45+\theta/2)).\]](http://latex.artofproblemsolving.com/1/1/1/111bf5c8ae549da2290126dbc5db67801c11d351.png)
Notice that if 
, then 
, and also 
, giving 
. Hence, the sequence of 
s is increasing, yet bounded above by 
, meaning this sequence converges. Then, we need 
, giving 
, so ![\[\lim_{n\to\infty}x_n = 2\cos(90) = 0, \lim_{n\to\infty}y_n = 2\sin(90) = 2.\]](http://latex.artofproblemsolving.com/f/1/3/f13b5c13bb8d2b0da98fcdcff612e5e7e11ecbde.png)
and 
, then we have![\[ x_{n+1} = 2\sin\left(\frac{\theta}{2}\right) \qquad \text{and} \qquad y_{n+1} = 2\cos\left(\frac{\theta}{2}\right) \]](http://latex.artofproblemsolving.com/e/0/f/e0f5042ea88adb069a37840afe5754a400a764d3.png)
Proof: Plugging in and into the recursive definitions, we're left with the system of equations
We can solve from the second equation that 
. Substituting this into the first equation and applying double angle formula, we have![\[ \cos\left(\frac{\theta}{2}\right)y_{n+1} = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) \]](http://latex.artofproblemsolving.com/a/7/0/a70e454e7addee7fe9c951450cb0d8d407df8074.png)
which simplifies to get us 
. 
and 
,
![$f:\mathbb (0;1] \to\mathbb R$](http://latex.artofproblemsolving.com/7/9/9/799657d25f079cf2f55cd420ee41aa57bebc68a9.png)
such that![$$f(x)+f(1-x)=0\; \forall x \in (0;1] $$](http://latex.artofproblemsolving.com/7/d/e/7de74760f30216f280a168cd1e8c5737cb6bff0c.png)
?
and 
and 
(where all angle measures are in degrees) satisfy 
and both of the recurrence relations. As a result, as 
and 
seem very suspicious-looking, so we might consider using trig. After bit of computation without using trig, it is apparent that it is really difficult to find a limit without it. Hence, we conclude to use trig. We see that 
,
portion. This can either be through the tangent formulas or the sine cos formulas. Trying out the tangent formula just doesn't make much sense because then the numerators and denominators don't match up. OK... but we doesn't make the 
,
for some 
.
.
is a geometric sequence with factor 
.
and see the trig the entire problem falls apart. consider 
.
.
We get that 
This finishes. They are 1, 0 respectively.
,
,
,
.
and 
.
.
and 
and we are ready.