Mathematical Reflections 2012, Issue 6 - Problem O249

Problem:
Find all triples $(x,y,z)$ of positive integers such that $$\dfrac{x}{y}+\dfrac{y}{z+1}+\dfrac{z}{x}=\dfrac{5}{2}.$$

Proposed by Titu Andreescu.

Solution:
By the AM-GM Inequality it must be $$3\sqrt[3]{\dfrac{z}{z+1}} \leq \dfrac{x}{y}+\dfrac{y}{z+1}+\dfrac{z}{x}=\dfrac{5}{2},$$ i.e. $\dfrac{z}{z+1} \leq \dfrac{125}{216},$ which gives $z=1$.
Using the AM-GM Inequality once again we obtain
$$2\sqrt{\dfrac{x}{2}} \leq \dfrac{x}{y}+\dfrac{y}{2}+\dfrac{1}{x}=\dfrac{5}{2},$$
i.e. $\dfrac{x}{2} \leq \dfrac{25}{16}$, which gives $x=1,2,3$.

(i) If $x=1$, then $\dfrac{1}{y}+\dfrac{y}{2}+1=\dfrac{5}{2}$ and solving for $y$ we get $y=1,2$.
(ii) If $x=2$, then $\dfrac{2}{y}+\dfrac{y}{2}+\dfrac{1}{2}=\dfrac{5}{2}$ and solving for $y$ we get $y=2$.
(iii) If $x=3$, then $\dfrac{3}{y}+\dfrac{y}{2}+\dfrac{1}{3}=\dfrac{5}{2}$, which gives no solution.

Therefore, all the triples of positive integers which satisfy the given equation are $$(1,1,1),(1,2,1),(2,2,1).$$

Mathematical Reflections 2012, Issue 6 - Problem O247

Problem:
Solve in positive integers the equation $$xy+yz+zx-5\sqrt{x^2+y^2+z^2}=1.$$

Proposed by Titu Andreescu.

Solution:
Clearly, $xy+yz+zx \geq 3$. Rewriting the equation in the form $$(xy+yz+zx-1)^2=25(x^2+y^2+z^2),$$ we put $xy+yz+zx=5t+1, t > 0$, so that $x^2+y^2+z^2=t^2$. Then $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=t^2+2(5t+1)=(t+5)^2-23,$$ which gives $$(t+5-x-y-z)(t+5+x+y+z)=23.$$
Since $x,y,z,t$ are positive integers, it must be
$$\left\{\begin{array}{rcl} t+5-x-y-z & = & 1 \\ t+5+x+y+z & = & 23. \end{array} \right.$$
Summing up the two equations, we get $t=7$. So, $x+y+z=11$ and $xy+yz+zx=36$. Suppose without loss of generality that $x \leq y \leq z$. Clearly, $x\leq3$, so we have three cases.

(i) $x=1$. Therefore, $y+z=10, yz=26$, so there are no solution.
(ii) $x=2$. Therefore, $y+z=9, yz=18$, so $y=3, z=6$.
(iii) $x=3$. Therefore, $y+z=8, yz=12$, so there are no solution.

In conclusion, all the positive integer solutions are $$(2,3,6), (2,6,3), (3,2,6), (3,6,2), (6,2,3), (6,3,2).$$

Mathematical Reflections 2012, Issue 6 - Problem U249

Problem:
Let $(a_n)_{n \geq 1}$ be a decreasing sequence of positive numbers. Let $$s_n=a_1+a_2+\ldots+a_n,$$ and $$b_n=\dfrac{1}{a_{n+1}}-\dfrac{1}{a_n},$$ for all $n \geq 1$. Prove that if $(s_n)_{n \geq 1}$ is convergent, then $(b_n)_{n \geq 1}$ is unbounded.

Proposed by Bogdan Enescu.

Solution:
Since $(s_n)_{n \geq 1}$ is convergent, then $\lim_{n \to \infty} a_n=0$. Suppose that $(b_n)_{n \geq 1}$ is bounded. Then there exists $M \in \mathbb{R}^+$ such that $b_n \leq M$ for all $n \geq 1$. By the Problem 2.5.12 at page 97 of Radulescu, Andreescu - Problems In Real Analysis, the sequence $(a_n)_{n \geq 1}$ converges to zero if and only if the series $$\sum_{n=1}^\infty \left(1-a_{n+1}/a_n\right)=\sum_{n=1}^\infty a_{n+1}b_n$$ diverges. But
$$\sum_{n=1}^\infty a_{n+1}b_n \leq \sum_{n=1}^\infty a_nb_n \leq M\sum_{n=1}^\infty a_n < \infty,$$ contradiction.

Mathematical Reflections 2012, Issue 6 - Problem U247

Problem:
Let $a$ be a real number greater than $1$. Evaluate
$$\dfrac{1}{a^2-a+1}-\dfrac{2a}{a^4-a^2+1}+\dfrac{4a^3}{a^8-a^4+1}-\dfrac{8a^7}{a^{16}-a^{8}+1}+\ldots.$$

Proposed by Titu Andreescu.

Solution:
Let $$S_n=\sum_{k=1}^n (-1)^{k-1}\dfrac{2^{k-1}a^{2^{k-1}-1}}{a^{2^k}-a^{2^{k-1}}+1}.$$ Since
$$\begin{array}{rcr} \dfrac{1}{a^2-a+1}-\dfrac{1}{a^2+a+1}&=&\dfrac{2a}{a^4+a^2+1} \\  -\dfrac{2a}{a^4-a^2+1}+\dfrac{2a}{a^4+a^2+1}&=&-\dfrac{4a^3}{a^8+a^4+1} \\ \vdots & \vdots & \vdots \\ (-1)^{n-1} \dfrac{2^{n-1}a^{2^{n-1}-1}}{a^{2^n}-a^{2^{n-1}}+1}+(-1)^n \dfrac{2^{n-1}a^{2^{n-1}-1}}{a^{2^n}+a^{2^{n-1}}+1} & = & (-1)^{n+1} \dfrac{2^na^{2^n-1}}{a^{2^{n+1}}+a^{2^n}+1}. \end{array}$$
summing up the two columns, we get $$S_n-\dfrac{1}{a^2+a+1}=(-1)^{n+1} \dfrac{2^na^{2^n-1}}{a^{2^{n+1}}+a^{2^n}+1}.$$
Now, $$0 \leq \left|(-1)^{n+1}\dfrac{2^na^{2^n-1}}{a^{2^{n+1}}+a^{2^n}+1}\right| \leq \dfrac{2^na^{2^n}}{a^{2^{n+1}}}=\dfrac{2^n}{a^{2^n}}$$ and  $\dfrac{2^n}{a^{2^n}} \to 0$ as $n \to \infty$. So,
$$\sum_{k=1}^\infty (-1)^{k-1}\dfrac{2^{k-1}a^{2^{k-1}-1}}{a^{2^k}-a^{2^{k-1}}+1}=\lim_{n \to \infty} S_n = \dfrac{1}{a^2+a+1}.$$

Mathematical Reflections 2012, Issue 6 - Problem S251

Problem:
Find all triples $(x,y,z)$ of positive real numbers for which there is a positive real number $t$ such that the following inequalities hold simultaneously:
$$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+t\leq4, \qquad x^2+y^2+z^2+\dfrac{2}{t}\leq5.$$

Proposed by Titu Andreescu.

Solution:
From the two inequalities, it's easy to see that if there exists such a positive real number $t$, then $t$ satisfies
\begin{equation}\label{first}
\dfrac{2}{5}<t<4.                                                              (1)
\end{equation}
From the first inequality, using the AM-HM inequality we obtain $$\dfrac{9}{4-t} \leq \dfrac{9}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \leq x+y+z.$$
Now, suppose without loss of generality that $x \geq y \geq z$. Then, $x^2 \geq y^2 \geq z^2$ and $1/x \leq 1/y \leq 1/z$.
By Chebyshev's Inequality, we have $$x+y+z \leq \dfrac{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)(x^2+y^2+z^2)}{3} \leq \dfrac{(5-2/t)(4-t)}{3}.$$ Hence, $$\dfrac{9}{4-t} \leq \dfrac{(5-2/t)(4-t)}{3}$$ and after simple calculations we get
$$5t^3-42t^2+69t-32 \geq 0 \iff (t-1)^2(5t-32) \geq 0,$$ which gives
\begin{equation}\label{second}
t=1, \qquad t \geq \dfrac{32}{5}.                                       (2)
\end{equation}
From $(1)$ and $(2)$, we get $t=1$, from which $x+y+z=1/x+1/y+1/z=3$. But this implies that the arithmetic mean and harmonic mean are equal, so $x=y=z=1$ and the only triple is $(1,1,1)$.

Mathematical Reflections 2012, Issue 6 - Problem S249

Problem:
Find the minimum of $2^x-4^x+6^x-8^x-9^x+12^x$ where $x$ is a positive real number.

Proposed by Titu Andreescu.

Solution:
We have $$\begin{array}{rcl}2^x-4^x+6^x-8^x-9^x+12^x&=&(4^x-3^x)(3^x-2^x)-[(4^x-3^x)+(3^x-2^x)]\\&=&(4^x-3^x-1)(3^x-2^x-1)-1. \end{array}$$
Let $f_n(x)=(n+1)^x-n^x-1$, where $n$ is a positive integer. $f_n(x)$ is increasing since
$$f'_n(x)=(n+1)^x\log(n+1)-n^x\log n > 0 \iff \left(1+\dfrac{1}{n}\right)^x>\dfrac{\log n}{\log (n+1)}$$ for all $x \in \mathbb{R}^+$. Moreover $f_n(1)=0$, so $f_n(x) \geq 0$ if $x \in [1,\infty)$ and $f_n(x) \leq 0$ if $x \in (0,1]$, for all $n \in \mathbb{N}^*$. Therefore, $$f_3(x)f_2(x)-1 \geq -1 \qquad \forall x \in \mathbb{R}^+$$ and the equality occurs if and only if $x=1$.

Mathematical Reflections 2012, Issue 6 - Problem S248

Problem:
Let $\mathcal{C}(O,R)$ be a circle and let $P$ be a point in its plane. Consider a pair of diametrically
opposite points $A$ and $B$ lying on $\mathcal{C}$. Prove that while points $A$ and $B$ vary on the
circumference of $\mathcal{C}$, the circumcircles of triangles $ABP$ pass through another fixed point.

Proposed by Ivan Borsenco.

Solution:
Let us consider another pair of diametrically opposite points $A'$ and $B'$ on $\mathcal{C}$. Then, $O$ is the midpoint of $AB$ and $A'B'$, so in the triangles $ABP$ and $A'B'P$ the points $O$ and $P$ are fixed, and this implies that the line $PO$ is fixed. Therefore, the line $PO$ intersects the circumcircle of the triangle $A'B'P$ in another point $Q$, which is fixed. By the arbitrarity of the pair of diametrically opposite points $A',B'$, we obtain that all the circumcircles of triangles $ABP$ pass through the point $Q$.

Mathematical Reflections 2012, Issue 6 - Problem S247

Problem:
Prove that for any positive integers $m$ and $n$, the number $8m^6 + 27m^3n^3 + 27n^6$ is
composite.

Proposed by Titu Andreescu.

Solution:
For the homogeneity of the expression, we set $t=m/n$. Then,
$$\begin{array}{rcl} 8t^6+27t^3+27 &= & 8t^6-27t^3+27+54t^3\\ &=& (2t^2)^3+(-3t)^3+3^3-(2t^2)(-3t)(3)\\&=&(2t^2-3t+3)(4t^4+6t^3+3t^2+9t+9). \end{array}$$ The second factor is clearly greater than $1$ and $2t^2-3t+3=2(t-1)^2+t+1>1$, so multiplying both sides of the equation by $n^6$ we obtain the conclusion.

Mathematical Reflections 2012, Issue 6 - Problem J251

Problem:
Let $a,b,c$ be positive real numbers such that $a \geq b \geq c$ and $b^2>ac$. Prove that
$$\dfrac{1}{a^2-bc}+\dfrac{1}{b^2-ca}+\dfrac{1}{c^2-ab}>0.$$

Proposed by Titu Andreescu.

Solution:
By AM-GM Inequality,
$$\dfrac{1}{a^2-bc}+\dfrac{1}{b^2-ca}+\dfrac{1}{c^2-ab} \geq \dfrac{3}{a^2+b^2+c^2-ab-bc-ca} > 0,$$
where the last inequality follows from the fact that $a,b,c$ cannot all be equal.

Mathematical Reflections 2012, Issue 6 - Problem J249

Problem:
Find the least prime $p>3$ that divides $3^q-4^q+1$ for all primes $q>3$.

Proposed by Titu Andreescu.

Solution:
Since $3^5-4^5+1=-780=-(2^2\cdot3\cdot5\cdot13)$, and $3^7-4^7+1$ is not divisible by $5$, we only have to show that $p=13$ divides $3^q-4^q+1$ for all primes $q>3$. Since $q$ is a prime, $q=6k \pm 1$ for some $k \in \mathbb{N}^*$. Moreover, $3^{6k} \equiv 1 \pmod{13}$ and $4^{6k} \equiv 1 \pmod{13}$, then
$$\begin{array}{rrrrrr} 3^{6k+1}-4^{6k+1}+1 & \equiv & 3-4+1 \equiv & 0 & \pmod{13} \\ 3^{6k-1}-4^{6k-1}+1 & \equiv & 9-10+1 \equiv & 0 & \pmod{13}, \end{array}$$
which gives the conclusion.

Mathematical Reflections 2012, Issue 6 - Problem J248

Problem:
Let $f:[1,\infty) \longrightarrow \mathbb{R}$ be defined by $f(x)=\dfrac{\{x\}^2}{\lfloor x \rfloor}$. Prove that $f(x+y) \leq f(x)+f(y)$, for any real numbers $x$ and $y$.

Proposed by Sorin Radulescu.

Solution:
Since $\{x+y\} \leq \{x\}+\{y\}$ and $\lfloor x \rfloor + \lfloor y \rfloor \leq \lfloor x+y \rfloor$ for all $x,y \in \mathbb{R}$, using the well known inequality $$\dfrac{a^2}{x}+\dfrac{b^2}{y} \geq \dfrac{(a+b)^2}{x+y} \qquad \forall a,b,x,y \in \mathbb{R}, x,y>0$$ we have $$f(x+y)=\dfrac{\{x+y\}^2}{\lfloor x+y \rfloor} \leq \dfrac{(\{x\}+\{y\})^2}{\lfloor x \rfloor + \lfloor y \rfloor} \leq \dfrac{\{x\}^2}{\lfloor x \rfloor} + \dfrac{\{y\}^2}{\lfloor y \rfloor}=f(x)+f(y)$$ for all $x,y \in [1,\infty)$.

Mathematical Reflections 2012, Issue 6 - Problem J247

Problem:
Let $a$ and $b$ be distinct zeros of the polynomial $x^3-2x+c$. Prove that $a^2(2a^2+4ab+3b^2)=3$ if and only if $b^2(3a^2+4ab+2b^2)=5$.

Proposed by Titu Andreescu.

Solution:

Let $\alpha$ be the other root of the polynomial $x^3-2x+c$. Then $\alpha^3-2\alpha+c=0$ and $$a+b=-\alpha, \qquad ab=\alpha^2-2.$$ Therefore, $$\begin{array}{rcl} a^2(2a^2+4ab+3b^2)+b^2(3a^2+4ab+2b^2) & = & 2(a+b)^2(a^2+b^2)+2a^2b^2 \\ & = & 2\alpha^2 [\alpha^2-2(\alpha^2-2)]+2(\alpha^2-2)^2 \\ &=& 2[\alpha^2-(\alpha^2-2)]^2 \\ &=& 8, \end{array}$$ and the statement follows.