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    (Original post by joostan)
    Problem 530: */**
    Let a,b,c,d>0 be arbitrary constants.
    Find all possible values of:
    S=\dfrac{a}{a+b+d}+\dfrac{b}{a+b  +c}+\dfrac{c}{b+c+d}+\dfrac{d}{a  +c+d}
    Solution 530
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    We prove that  \dfrac{a}{a+b+d}+\dfrac{b}{a+b+c  }+\dfrac{c}{b+c+d}+\dfrac{d}{a+c  +d}<2 .

    Lemma: If  0<m<n then  \frac{m}{n}<\frac{m+a}{n+a} for some positive  a .
    Proof: Multiply out.

    We note that for each of the four terms in  S , the numerator is smaller than the denominator, so we can have
    S< \dfrac{a+c}{a+b+c+d}+ \dfrac{b+d}{a+b+c+d}+ \dfrac{c+a}{a+b+c+d}+ \dfrac{d+b}{a+b+c+d}=2 .

    This is achieved as we let  b,d \rightarrow 0 .

    We now prove that  \dfrac{a}{a+b+d}+ \dfrac{b}{a+b+c}+ \dfrac{c}{b+c+d}+ \dfrac{d}{a+c+d} >1 .
    By CS,
    S(a(a+b+d)+b(a+b+c)+c(b+c+d)+d(a  +c+d)\geq (a+b+c+d)^2
    Then
     S\geq \frac{a^2+b^2+c^2+d^2+2ab+2bc+2c  d+2da+2ac+2bd}{a^2+b^2+c^2+d^2+2  ab+2bc+2cd+2da}= 1+ \frac{2ac+2bd}{a^2+b^2+c^2+d^2+2  ab+2bc+2cd+2da} > 1.

    This is achieved as we let  a,b \rightarrow 0

    Thus  1<S<2

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    Problem 531: */**

    Give a geometrical argument to show that, if x^2+y^2=1, then

    \displaystyle \int_{\frac{\sqrt{2}}{2}}^{ \frac{\sqrt{2+\sqrt{2}}}{2} } \frac{1}{\sqrt{1-x^2}} \ dx = \frac{\pi}{8}
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    (Original post by Zacken)
    Is the denominator of b/etc correct?
    Nope
    Will edit, thanks
    Though someone seems to have corrected me to solve it.
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    Problem 532 */**

    Define a sequence by  u_1 = 1,     u_2 = 1,  u_{n+1} = u_n + u_{n-1} for n = 3, 4, 5,...

    Prove that the simultaneous equations

     u_{k} x + u_{k+1} y + u_{k+2} z = a
     u_{m} x + u_{m+1} y + u_{m+2} z = b
     u_{n} x + u_{n+1} y + u_{n+2} z = c

    Have no unique solutions.

    Comment on Problem 529

    Nice joostan. In addition to "brute forcing" it there is a trick, which I wondered if anybody would get.
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    Solution 525

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    Observe that

    \displaystyle \dfrac{1}{n} = \dfrac{1}{x} - \left(\dfrac{1}{x} - \dfrac{1}{n}\right) = \dfrac{1}{x} + \int_{n}^{x} \dfrac{1}{t^2} \ dt

    So, by partial summation, we have

     \displaystyle \sum_{n \leq x} \dfrac{1}{n} = \dfrac{1}{x} \sum_{n \leq x} 1 + \int_{1}^{x} \left(\sum_{n\leq t} 1 \right) \dfrac{1}{t^2} \ dt

    \displaystyle = \dfrac{[x]}{x} + \int_{1}^{x} \dfrac{[t]}{t^2} \ dt

    Now we introduce the subscript f to denote "fractional part of"

    \displaystyle = \dfrac{x - x_f}{x} + \int_{1}^{x} \dfrac{t - t_f}{t^2} \ dt

     = \displaystyle 1 - \dfrac{x_f}{x} + \log x - \int_{1}^{\infty} \dfrac{t_f}{t^2} \ dt + \int_{x}^{\infty} \dfrac{t_f}{t^2} \ dt

    Thus we see that

    \displaystyle  E = 1 - \int_{1}^{\infty} \dfrac{t_f}{t^2} \ dt

    since the final integral can be bounded as

    \displaystyle \int_{x}^{\infty} \dfrac{t_f}{t^2} \ dt \leq \int_{x}^{\infty} \dfrac{1}{t^2} \ dt = \dfrac{1}{x}

    Hence we get the final result with an error term of O(x^{-1}), which the second term also contributes to.

    By the way, E is the Euler-Mascheroni constant.

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    Solution 532

    Without loss of generality, let  k < m < n .
    The inequality is strict as they cannot be equal as that would mean two of the simultaneous equations are equivalent so we would have no unique solution.

    We aim to show that a linear combination of the first simultaneous equation and the second gives you the third simultaneous equation, as if that is the case then we only have 2 equations and 3 variables so there would be no unique solution.

    There is such a linear combination if there exists  d,e such that
     du_k + eu_m = u_n \hspace{3mm} du_{k+1} + eu_{m+1} = u_{n+1}
    as by adding those two equations you get
     du_{k+2} + eu_{m+2} = u_{n+2}

    By using the idea of adding equations again, we can see by repeated addition of the equations
     eu_{m-k} = u_{n-k} \hspace{3mm} d + eu_{m-k+1} = u_{n-k+1}
    we get the condition for the linear combination.

     e = \frac{u_{n-k}}{u_{m-k}} \hspace{3mm} d = u_{n-k+1} - \frac{u_{n-k}}{u_{m-k}}u_{m-k+1}
    and so we have shown the linear combination exists which in turn means there is no unique solution to the set of simultaneous equations.
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    (Original post by 16Characters....)
    Problem 532 */**


    Comment on Problem 529

    Nice joostan. In addition to "brute forcing" it there is a trick, which I wondered if anybody would get.
    Well the other way I saw you could do it is:
    I=\displaystyle\int e^xsech(x) \ dx

J=\displaystyle\int e^{-x}sech(x) \ dx

\Rightarrow \dfrac{I+J}{2} = \displaystyle\int \ dx

\dfrac{I-J}{2} = \displaystyle\int \tanh(x) \ dx
    Hence:
    I=\displaystyle\int 1+\tanh(x) \ dx=x+\ln(\cosh(x))+\mathcal{C}

J=\displaystyle\int 1-\tanh(x) \ dx=x-\ln(\cosh(x))+\mathcal{C}
    But it didn't really seem worth it.
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    Problem 533 *

    Consider a rectangle with sides of length (x, y, x, y). Find a quadratic equation satisfied by x and y and hence prove that

     S \geq 2\sqrt A

    Where S and A are the semiperimeter and area of the rectangle respectively.
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    (Original post by 16Characters....)
    Problem 533 *

    Consider a rectangle with sides of length (x, y, x, y). Find a quadratic equation satisfied by x and y and hence prove that

     S \geq 2\sqrt A

    Where S and A are the semiperimeter and area of the rectangle respectively.
    Solution 533

    Not really sure what the quadratic I want to find is, but we have
     S=x+y
     A=xy
    Then
     S^2=x^2+y^2+2xy\geq 4xy=4A by AM-GM so
     S\geq 2 \sqrt{A}
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    (Original post by Renzhi10122)
    Solution 533
    I only included about finding a quadratic as a starting point since that provides a C1-level method without the need for AM-GM or anything fancy like that.
    Spoiler:
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    The quadratic in question is  w^2 - Sw + A = 0 which is easily obtained by writing  (w-x)(w-y) = 0 is a quadratic satisfied by x,y and expanding (or alternatively using properties of roots from FPx modules).

    The discriminant then gives the desired inequality, once it is remembered S, A are both non-negative so  S^2 \geq 4A \implies S \geq 2\sqrt A
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    (Original post by 16Characters....)
    I only included about finding a quadratic as a starting point since that provides a C1-level method without the need for AM-GM or anything fancy like that.
    Spoiler:
    Show

    The quadratic in question is  w^2 - Sw + A = 0 which is easily obtained by writing  (w-x)(w-y) = 0 is a quadratic satisfied by x,y and expanding (or alternatively using properties of roots from FPx modules).

    The discriminant then gives the desired inequality, once it is remembered S, A are both non-negative so  S^2 \geq 4A \implies S \geq 2\sqrt A
    Ah, fair enough. I could have just said that  (x-y)^2\geq 0 but it's quicker to just say by AMGM.
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    (Original post by Renzhi10122)
    Ah, fair enough. I could have just said that  (x-y)^2/geq 0 but it's quicker to just say by AMGM.
    Yes I realised that also when reading your solution. I hadn't thought of that whilst writing the question :-)
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    Problem 534

    Prove that, \forall n \in \mathbb{N}, if n is odd, then f(x) = x^n is a strictly increasing function on \mathbb{R}
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    (Original post by Indeterminate)
    Problem 534
    Can I assume that f is continuous or will I have to prove that also?
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    (Original post by 16Characters....)
    Can I assume that f is continuous or will I have to prove that also?
    Yes, you may assume that f is continuous (that's a triviality imo).
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    Solution 534

    Not an undergraduate so never formally studied any analysis-y stuff so please tell me if this is insufficient.
    Spoiler:
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    Since  f(x) = x^n with n an odd natural number then  f'(x) = nx^{n-1} . Since n is odd then n-1 is even so can be written as  n - 1 = 2r for some non-negative integer r. Hence

     f'(x) = n(x^r)^2
    For real x,  x^r is real. The square of a positive real number is always postive hence  f'(x) \geq 0 with equality occuring only when  x = 0 . Hence for  x \neq 0,  f'(x) > 0 is strictly increasing since f is continuous.

    When  x = 0 ,  f(x) = 0 . For x infinitesimally larger then 0  f(x) > 0 = f(0) so the function is still strictly increasing at this point.

    Hence f is strictly increasing over the reals.
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    Problem 535

    Find all positive integers n such that n divides (n-1)!

    Posted from TSR Mobile
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    Problem 536

    Can an irrational number to the power of an irrational number be rational?

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    Solution 536:

    We want to find irrational a and b such that a^b is rational. Consider \displaystyle \sqrt{2}^{\sqrt{2}}, if that is rational, we are done with a = b = \sqrt{2}.

    If not, take \displaystyle a = \sqrt{2}^{\sqrt{2}} and b = \sqrt{2}, then a^b = \sqrt{2}^{\sqrt{2} \cdot \sqrt{2}} = \sqrt{2}^2 = 2 and we are done.
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    (Original post by Krollo)
    Problem 535

    Find all positive integers n such that n divides (n-1)!

    Posted from TSR Mobile
    Solution 535

    We find all integers n that do not divide (n-1)!
    Suppose the highest power of a prime  p that divides  n is  a . Then there are at least  p^{a-1}-1 multiples of  p that are less than  n .
    Suppose that  a\geq2 and  p>2 then we claim that  a\leq p^{a-1}-1 . This is clearly true for  a=2 and the induction is trivial. If  p=2 then we have  a\leq 2^{a-1}-1 for  a\geq 3 . If  a=2,p=2 , then the only  n for which  (n-1)! is not divisible by 4 and  4|n is  n=4 . Thus,  p^a divides  (n-1)! for  n \neq 4 .We will now assume that  n\neq 4 .
    Thus if the power of any prime in the prime factorisation of  n is not 1, then  (n-1)! is divisible by this power of the prime. We may now suppose that  n=p_1p_2...p_k for  p_1<p_2<...<p_k .
    Suppose  k\geq2 , then there exist at least  \frac{n}{p_i}>1 multiples of  p_i that are less than  n . Thus  p_i divides  (n-1)! , and it follows that  n divides  (n-1)! . If  k=1 , then clearly  n does not divide  (n-1)! , so our solutions are all the integers that are not prime or 4.
 
 
 
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