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Convergent sequence

Let \{x_n\}_{n \in \mathbb{N}} be a sequence of positive real numbers such that

(1)   \begin{equation*} x_{n+m} \leq x_n + x_m \quad , \quad  m, n \in \mathbb{N} \end{equation*}

Prove that \left\{\dfrac{x_n}{n} \right\}_{n \in \mathbb{N}} converges.


Fix m and let n \geq m. Then, there exist k, r such that n=km+r where 0\leq r <m. Thus,

    \[\frac {x_n}{n} = \frac {x_ {km+r}}{n}\leq \frac {kx_ {m}}{n} + \frac {x_{r}}{n}\]

Letting n \rightarrow +\infty it follows that

    \[\limsup \frac {x_n}{n} \leq \frac {x_ {m}}{m} + 0\]

Since this holds forall m it follows that \displaystyle \limsup \frac {x_n}{n} \leq \liminf \frac {x_ {m}}{m} and the result follows.

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Root inequality

Let a, b, c, d be positive real numbers satisfying the following equality

    \[\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} =4\]

Prove that

\displaystyle \sqrt[3]{\frac{a^3+b^3}{2}} + \sqrt[3]{\frac{b^3+c^3}{2}} + \sqrt[3]{\frac{c^3+d^3}{2}} + \sqrt[3]{\frac{d^3+a^3}{2}} \leq 2 \left ( a+b+c+d \right ) -4


We begin by stating a lemma:

Lemma: Let a, b be positive real numbers, then:

    \[\frac{a+b}{2} \leq \sqrt[3]{\frac{a^3+b^3}{2}} \leq \frac{a^2+b^2}{a+b}\]

Now, making use of the lemma we have that:

    \begin{align*} \sum \sqrt[3]{\frac{a^3+b^3}{2}} &\leq \sum \frac{a^2+b^2}{a+b} \\ &=\sum \left ( a+b \right )-\sum \frac{2ab}{a+b} \\ &= 2\left ( a+b+c+d \right )- 2\sum \frac{1}{\frac{1}{a}+\frac{1}{b}} \end{align*}

Making use of the Cauchy – Schwartz inequality we have that

    \begin{align*} \sum \frac{1}{\frac{1}{a}+\frac{1}{b}} & \geq \frac{\left ( 1+1+1+1 \right )^2}{2\left ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c}+ \frac{1}{d} \right )} \\ &=\frac{8}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}+ \frac{1}{d}} \\ &= 2 \end{align*}

The inequality now follows.

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Trigonometric identity

Let a, b, c be positive real numbers such that a+b+c=\pi. Prove the following trigonometric identities:

  1. \displaystyle \sin a + \sin b + \sin c = 4 \cos \frac{a}{2} \cos \frac{b}{2} \cos \frac{c}{2}
  2. \displaystyle \cos a + \cos b + \cos c = 1+4 \sin \frac{a}{2} \sin \frac{b}{2} \sin \frac{c}{2}


  1. We have successively:

        \begin{align*} \sin a + \sin b +\sin c &= 2 \sin \frac{a+b}{2} \cos \frac{a-b}{2} + \sin c \\ &=2 \sin \left ( \frac{\pi}{2} - \frac{c}{2} \right ) \cos \frac{a-b}{2} + 2 \sin \frac{c}{2} \cos \frac{c}{2} \\ &=2 \cos \frac{c}{2} \cos \frac{a-b}{2} + 2 \sin \frac{c}{2} \cos \frac{c}{2} \\ &=2 \cos \frac{c}{2} \left ( \cos \frac{a-b}{2} + \sin \frac{c}{2} \right ) \\ &= 2 \cos \frac{c}{2} \left ( \cos \frac{a-b}{2}+ \cos \frac{a+b}{2} \right ) \\ &= 2 \cos \frac{c}{2} \cdot 2 \cos \frac{a}{2} \cos \frac{b}{2} \\ &= 4 \cos \frac{a}{2} \cos \frac{b}{2} \cos \frac{c}{2} \end{align*}

  2. Similarly, we have successively:

        \begin{align*} \cos a + \cos b + \cos c &=2 \cos \frac{a+b}{2} \cos \frac{a- b}{2} + \cos c \\ &=2 \cos \left ( \frac{\pi}{2} - \frac{c}{2} \right ) \cos \frac{a-b}{2} + \cos c \\ &=2 \sin \frac{c}{2} \cos \frac{a-b}{2} + \left ( 1 - 2 \sin^2 \frac{c}{2} \right ) \\ &=1 + 2 \sin \frac{c}{2} \left (\cos \frac{a-b}{2} - \sin \frac{c}{2} \right ) \\ &=1 + 2 \sin \frac{c}{2} \left [ \cos \frac{a-b}{2} - \sin \left ( \frac{\pi}{2} -\frac{a+b}{2} \right ) \right ] \\ &=1 + 2 \sin \frac{c}{2} \left ( \cos \frac{a-b}{2} - \cos \frac{a+b}{2} \right ) \\ &=1 + 2 \sin \frac{c}{2} \cdot 2 \sin \frac{a}{2} \sin \frac{b}{2} \\ &=1 + 4 \sin \frac{a}{2} \sin \frac{b}{2} \sin \frac{c}{2} \end{align*}

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Number of solutions

Find the number of solutions of the equation:

    \[x_1+x_2+\cdots + x_{100} = 2017\]

in the positive integers.


We set y_n=x_n-1 \;\;, \;\; 1\leq n \leq 100. It suffices to find the number of solutions of the equation

(1)   \begin{equation*} y_1+y_2+ \cdots + y_n =1917 \end{equation*}

in the non negative numbers. We represent each sum z_1+\cdots+z_m of non negative integers with a sequence of z_1 dots (\bullet) followed by a vertical bar (\big \mid), after z_2 dots another one vertical bar etc, till we place the last z_m dots ( without the vertical bar at the end.) For example the sum 5+2+0+1=8 can be represented as

    \[\bullet \bullet \bullet  \bullet \bullet \mid \bullet \bullet \mid \mid \bullet\]


We note that every solution of (1) matches a sequence that has 1917 dots in total and 99 vertical bars. Conversely, every such sequence matches a solution of (1).

Thus, in total there are

    \[\binom{1917+99}{99} = \binom{2016}{99}\]


Comment: In general the equation


has \displaystyle  \binom{n-1}{m-1} solutions in the positive integers and \displaystyle \binom{n+m-1}{m-1} solutions in the non negative integers.


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Eulerian equality

We know that there are infinite Pythagorian triplets, that is numbers a, b, c such that

(1)   \begin{equation*} a^2 = b^2 +c^2 \end{equation*}

Let us investigate if there exist triplets such that

(2)   \begin{equation*} \varphi \left( a^2 \right) = \varphi \left( b^2 \right) + \varphi \left( c^2 \right) \end{equation*}

where \varphi denotes the Euler’s totient function.


Indeed, there are infinite triplets such that (2) is satisfied. For example noticing that

    \[\varphi \left(4^2 \right) + \varphi \left(6^2 \right) = 8 + 12 = 20 = \varphi \left(5^2 \right)\]

we deduce that for each natural N such that (N, 30) =1 we have

    \[\varphi \left((4N)^2 \right) + \varphi \left((6N)^2 \right) = 20\varphi \left(N^2\right) = \varphi \left((5N)^2 \right)\]

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