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    (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=8, we have f(1,7,8)< \frac{3}{2} = f(1,1,1). Hence f(a,b,c) \ge \frac{3}{2a} for all a,b,c > 0 is clearly not true. I have not tried f(a,b,c) \ge \frac{3}{2c}, but even if it is true, then you have to prove that RHS \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 \in S, f(x) \ge g(x). Then, you are claiming that f(x) \ge \max(g(x)_{x\in S}), which is generally not true.

    Spoiler:
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    I recommend using some well-known inequalities, in which case, the inequality is proven in less than a minute.
    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!
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    Solution 73

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

    Problem 74 hint:
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    Imprimis, we can apply C-S to \displaystyle \frac{a}{(b+c)^{2}}+ \frac{b}{(c+a)^{2}}+ \frac{c}{(a+b)^{2}}. The rest follows from AM-GM.
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    (Original post by Mladenov)
    Solution 73

    I assume n \ge 3. Let S be a finite multi-set of n positive integers. If all elements of S are greater than 1, we know that the product of all the elements of S is greater than or equal than their sum.
    Suppose that we have solution with exactly k components greater than 1. Thus the remaining n-k components are equal to 1. Now, the sum of the elements, which are greater than 1, is less than or equal to the product; the sum of the n-k 1's is less than or equal to n-1. It follows that k=1.
    As we are counting ordered sets, we have n(m-1)+1 solutions in positive integers.
    The number of non-negative integer solutions of the equation x_{1}+..+x_{n-1} = n-1 is \dbinom{2n-3}{n-1}. Hence, the number of solutions in non-negative integers is 2\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!
    Btw, I'm making some progress with your problem now using
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    c-s in Engel Form
    which I wasn't familiar with until now (hopefully just a few algebraic tricks away from solution!)
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    Solution 61

    Assume f continuous (perhaps this was not necessary).

    Let \mathcal{P}=(0,1),\;\mathcal{Q}=  (1,\infty), noting that f(1)=1. If f(s)=1 for some s\in\mathcal{P}, then for all y,\;f(s^y)=1 \Rightarrow f(r)=1 over \mathcal{P}. However, then for y\in \mathcal{P}:\;f(x^y)=f(x) \Rightarrow f=\mathcal{C}, and hence f=1 over \mathbb{R}^+. Similarly if f(s)=1 for some s\in\mathcal{Q} then f=1 over \mathbb{R}^+. In a similar vein, suppose there exist distinct x_0,y_0\in \mathcal{P} such that f(x_0)=f(y_0)\neq 1 then f(x_0)^{f(r)}=f(y_0)^{f(r)} \Rightarrow f(x_0^r)=f(y_0^r) which implies f(x)=\mathcal{C} and since \mathcal{C}=\mathcal{C}^{ \mathcal{C}}\Rightarrow \mathcal{C}=\pm 1 we have f=1 over \mathcal{P}, which, as previously demonstrated, implies f=1 over \mathbb{R}^+. Similarly for the case x_0,x_1\in\mathcal{Q}. Hence if f is not injective it is 1 everywhere.

    Naturally, we now assume f is injective. Observe that 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(z^{y^n})=f(z)^{f(y)^n}. Combining these facts, f(n)=n for positive integers n. Hence we have f(x^n)=f(x)^n. Now suppose that f(x)>x over (a,b)\subset\mathcal{Q}, then by the preceding statement, f(x^n)>x^n over (a^n,b^n) and hence the inequality is satisfied over arbitrarily large intervals, which contradicts f(n)=n. Similarly if f(x)<x over any subinterval of \mathcal{Q}. Hence we must have f(x)=x over \mathcal{Q}. Now assuming f(x)>x over \mathcal{P}, our preceding proposition would imply that for y\in\mathcal{Q} and x\in\mathcal{P}:\; y^x=f(y^x)<f(y)^{f(x)}=f(y^x) which is absurd. Similarly if f(x)<x over \mathcal{P}. Hence f(x)=x over \mathcal{P} as well.

    Therefore f(x)=1 or f(x)=x.
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    (Original post by Lord of the Flies)
    ...
    Splendid solution.
    You are not using continuity, so it is not necessary.

    Another approach:
    Spoiler:
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    Clearly, f \equiv 1 is a solution. Suppose that there is at least one z such that f(z) \not= 1. We have f(xy)=f(x)f(y) and f(x+y)=f(x)+f(y) for all x,y\in \mathbb{R^{+}}. Since f is bounded below and satisfies f(x+y)=f(x)+f(y), it follows that f(x)=cx, c\in \mathbb{R^{+}}. Note that c^{2}xy=cxy implies c=1. Hence all the solutions are f \equiv 1 and f(x)=x.
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    (Still can't do problem 74 (fml!) so thought "if you can't beat them, join them")

    Problem 75


    Show that  \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 .
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    Solution 75

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

    We have \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 \displaystyle 2(x^{2}+y^{2}+z^{2})^{2} = 2\sum_{cyc} x^{4}+ 4\sum_{cyc} x^{2}y^{2}, from
    \displaystyle \sum_{cyc} x^{4} \ge \sum_{cyc} x^{3}y (AM-GM), and
    \displaystyle \sum_{cyc} (x^{4}+x^{2}y^{2}) \ge \sum_{cyc} 2x^{3}y (obvious), we obtain
    \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:
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    This is indeed nice application of Engel's form.
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    Use similar technique to prove 74. Without the substitution of course.
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    (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
    Geometry is wonderful, if you can do it
    Geometry is also taught extremely badly in most countries.
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    (Original post by Mladenov)
    \displaystyle \sum_{cyc} x^{4} \ge \sum_{cyc} x^{3}y (AM-GM), and
    \displaystyle \sum_{cyc} (x^{4}+x^{2}y^{2}) \ge \sum_{cyc} 2x^{3}y (obvious)
    I was wondering if you'd use homogenisation!
    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 \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:
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    I stopped trying it yesterday but I might have another go tomorrow. What I've got so far is:
    Spoiler:
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    By using normalisation I deduced that abc=1 can be used WLOG. I then applied engel's form to get  \frac{( \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} )^2}{ab+bc+ca} and I've fiddled around with it but haven't been able to find a way to nicely either algebraically combine it with the other art of the LHS, apply C-Z, rearrangement or AM-GM with the other part of the LHS, or find a way to apply one of these such a equalities to the expression itself such that it is greater than or equal to the difference of the RHS - *the other term of the LHS*. I've managed to get a few fairly simple expressions that the LHS is greater than but haven't quite hit the nail on the head.
    If I'm miles off, a hint would be helpful, but if I'm along the right lines then just give me a nod
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    (Original post by Llewellyn)
    Geometry is wonderful, if you can do it
    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.
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    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

     \displaystyle(a^2+b^2)^2 \geq (a+b+c)(a+b-c)(b+c-a)(c+a-b)
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    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 x^2 + y and x+y^2, and these numbers are distinct.

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

    Another approach:
    Spoiler:
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    Clearly, f \equiv 1 is a solution. Suppose that there is at least one z such that f(z) \not= 1. We have f(xy)=f(x)f(y) and f(x+y)=f(x)+f(y) for all x,y\in \mathbb{R^{+}}. Since f is bounded below and satisfies f(x+y)=f(x)+f(y), it follows that f(x)=cx, c\in \mathbb{R^{+}}. Note that c^{2}xy=cxy implies c=1. Hence all the solutions are f \equiv 1 and f(x)=x.
    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



    I should really be working, mais bon.

    Solution 52

    Note that since (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 (a_{n},a_{n-1})|a_{n-2} which then implies (a_{n},a_{n-1})|(a_{n-1},a_{n-2}) and inductively (a_n,a_{n-1})|(a_m,a_{m-1}) for any m\leq n. Hence there must be some n_0 such that n\geq n_0\Rightarrow (a_{n},a_{n-1})=\mathcal{D}. Then clearly for all n\geq n_0 we must have a_n=\mathcal{D} q_n where (q_i,q_j)=1 for any i,j\geq n_0. But since (a_n) is bounded (q_n) is bounded as well, and hence we require q_{n\geq N}=1 for some N\geq n_0. Thus a_{n\geq N}=\mathcal{D}, which, if plugged into the definition of (a_n) immediately gives a_{n\geq N}=2. Now we work backwards to show that all preceding terms are also 2. Obviously (a_N,a_{N-1})\in \{1,2\}, and given that 2(a_N,a_{N-1})=2+a_{N-1} we clearly cannot have (a_N,a_{N-1})=1 and hence it follows that a_{N-1}=2 etc. down to a_1=2.

    Therefore the only possible sequence is a_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\geq y\geq z and x^3\geq y^3\geq z^3. The second is just \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:

    \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 \displaystyle \lambda(x^2+y^2+z^2)^2\geq 3\lambda\sum_{\text{cyc}}x^3y and \displaystyle \mu(x^2+y^2+z^2)^2\geq 3\mu\sum_{\text{cyc}}x^2y^2 we get:
    \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 \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)=6x^7e^{x^2}\sin^2 (x^{1000}) find f^{(2013)}(0)-f^{(2012)}(0)
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    (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 x^2 + y and x+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.
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    (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)=6x^7e^{x^2}\sin^2 (x^{1000}) find 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.
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    (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.
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    (Original post by Jkn)
    ..
    AM-GM - \displaystyle \sum_{cyc} x^{4} \ge \sum_{cyc} x^{2}y^{2} and \displaystyle \sum_{cyc} (x^{4}+x^{2}y^{2}) \ge 2\sum_{cyc} x^{3}y, imply \displaystyle \sum_{cyc} x^{4} \ge \sum_{cyc} x^{3}y.
    Problem 74 hint
    Spoiler:
    Show
    Rewrite \displaystyle \frac{a}{(b+c)^{2}}+ \frac{b}{(c+a)^{2}}+ \frac{c}{(a+b)^{2}}= \sum_{cyc} \frac{a^{4}}{a^{3}(b+c)}. Now apply C-S to this sum. Then remove the denominators and apply AM-GM properly.


    (Original post by Lord of the Flies)
    Solution 52...
    C'est ça!

    Solution 76

    We can suppose a+b-c > 0, a+c-b >0, and b+c-a >0. AM-GM and C-S yield (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,d such that
    a^{2}+b=c^{2}+d;
    a+b^{2}=c+d^{2}.
    (a-c)(a+c)=d-b, a-c=d^{2}-b^{2}. Thus, a+c=d+b= \pm1. The case -1 implies c=-1-a, d=-1-b. So, a^{2}+b=c^{2}+d= 1+2a+a^{2}-1-b, a=b-contradiction. Let a+c=b+d=1. We then have a-c=d-b and therefore a=d, b=c. Consequently, a^{2}+b=d^{2}+c=a+b^{2}-contradiction. Hence our person (I do not remember his name ) has to dig in only one place.

    Solution 78

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

    \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 \displaystyle \lambda(x^2+y^2+z^2)^2\geq 3\lambda\sum_{\text{cyc}}x^3y and \displaystyle \mu(x^2+y^2+z^2)^2\geq 3\mu\sum_{\text{cyc}}x^2y^2 we get:
    \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
    Problem 78 (a classic!)

    Evaluate \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt  {\cdots}}}}
    Love this question! 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)
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    (Original post by Mladenov)
    AM-GM - \displaystyle \sum_{cyc} x^{4} \ge \sum_{cyc} x^{2}y^{2} and \displaystyle \sum_{cyc} (x^{4}+x^{2}y^{2}) \ge 2\sum_{cyc} x^{3}y, imply \displaystyle \sum_{cyc} x^{4} \ge \sum_{cyc} x^{3}y.
    But your solution says \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:
    Show
    Rewrite \displaystyle \frac{a}{(b+c)^{2}}+ \frac{b}{(c+a)^{2}}+ \frac{c}{(a+b)^{2}}= \sum_{cyc} \frac{a^{4}}{a^{3}(b+c)}. Now apply C-S to this sum. Then remove the denominators and apply AM-GM properly.
    ahhhhhh!
    C'est ça!

    Solution 76

    We can suppose a+b-c > 0, a+c-b >0, and b+c-a >0. AM-GM and C-S yield (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 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:
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    algebraic tricks
    than anything else (though you may find a different solution to me!)
    Solution 78

    Define f(x)=\sqrt{1+xf(x+1)}. It is quite well-known that f(x)=x+1. Hence \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt  {\cdots}}}}=3.
    There are several ways to solve this functional equation. We can bound f above and below; or we can introduce the function g(m,n) = \sqrt{1+(m-1)g(m+1,n)} and our expression is \displaystyle \lim_{n \to \infty} g(3,n); finally, we can consider the general equation \displaystyle f(x) = (g(x)+x^{\alpha}f(x+a))^{\frac{1  }{\beta}}, where x,a, \alpha \ge 0, \beta >1, and g : \mathbb{R^{+}} \cup 0 \to \mathbb{R^{+}} \cup 0, \inf g > 0, and solve it iteratively.
    Dayyyyyum son you post fast!
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    (Original post by Jkn)
    But your solution says \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 x^{3}y. From AM-GM we have \displaystyle \sum_{cyc} x^{4} \ge \sum_{cyc} x^{2}y^{2} and \displaystyle \sum_{cyc} (x^{4}+x^{2}y^{2}) \ge 2\sum_{cyc} x^{3}y, the second is also trivially true, and hence \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 \le a+b).

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

    Solution 79

    As x\to 0, from Taylor's series we have \sin (x) = x+\mathcal{O}(x^{3}), e^{x}=1+x+\frac{1}{2}x+\frac{1}{  6}x^{3}+ \mathcal{O}(x^{4}).
    Thus, 6x^{7}e^{x^{2}}\sin (x^{1000})= 6x^{7}(x^{1000}+\mathcal{O}(x^{2  000}))^{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!


    Problem 80

    Let (a_{n})_{n\ge1} be an increasing sequence of positive integers. Suppose also that \displaystyle \lim_{n\to \infty} \frac{a_{n}}{n}=0. Then, the sequence \displaystyle b_{n} = \frac{n}{a_{n}} contains all positive integers.
 
 
 
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