Mathematical Reflections 2016, Issue 5 - Problem O387

Problem:
Are there integers $n$ for which $$3^{6n-3}+3^{3n-1}+1$$ is a perfect cube?

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Clearly if $n \leq 0$, the number $3^{6n-3}+3^{3n-1}+1$ is never a perfect cube. Let $n>0$. Then,
$$(3^{2n-1})^3=3^{6n-3} < 3^{6n-3}+3^{3n-1}+1 < 3^{6n-3}+3^{4n-1}+3^{2n}+1=(3^{2n-1}+1)^3.$$
So, there are no integers $n$ such that $3^{6n-3}+3^{3n-1}+1$ is a perfect cube.

Mathematical Reflections 2016, Issue 5 - Problem O386

Problem:
Find all pairs $(m,n)$ of positive integers such that $3^m-2^n$ is a perfect square.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Let $n \in \mathbb{N}$ such that
$$3^x-2^y=n^2 \qquad (1).$$
We have three cases.

(i) If $y \geq 2$, then $3^x \equiv n^2 \pmod{4}$, so $x$ must be even, i.e. $x=2k$ for some $k \in \mathbb{N}^*$. Therefore, equation (1) becomes $$(3^k-n)(3^k+n)=2^y.$$ It follows that $3^k-n=2^a$ and $3^k+n=2^b$, where $a,b \in \mathbb{N}$ and $a+b=y$. Moreover, $a<b$ and adding these two equations, we get $$2\cdot3^k=2^a+2^b.$$ If $a=0$, the LHS is even and the RHS is odd, contradiction. If $a \geq 2$, we obtain $2\cdot3^k \equiv 0 \pmod{4}$, contradiction. So, $a=1$ and $b=y-1$ and we get $$3^k=1+2^{y-2}.$$ If $y=3$, we obtain $k=1$, i.e. $x=2$. If $y \geq 4$, then $3^k \equiv 1 \pmod{4}$, so $k$ is even and we can write $$(3^{\frac{k}{2}}-1)(3^{\frac{k}{2}}+1)=2^{y-2}.$$ Since $3^{\frac{k}{2}}+1$ and $3^{\frac{k}{2}}-1$ are powers of $2$ and their difference is $2$, we obtain $3^{\frac{k}{2}}+1=4$ and $3^{\frac{k}{2}}-1=2$, i.e. $k=2$, which gives $x=4$ and $y=5$. Therefore, we obtain the solutions $(x,y) \in \{(2,3),(4,5)\}$.

(ii) If $y=1$, then $3^x-2=n^2$. We have that $x$ must be odd, otherwise $n^2 \equiv -1 \pmod{4}$, contradiction. In the ring of integers $\mathbb{Z}[\sqrt{-2}]$, we have $$3^x=(n-\sqrt{-2})(n+\sqrt{-2}).$$
Let $d=(n-\sqrt{-2},n+\sqrt{-2})$. Clearly, $d \ | \ 2\sqrt{-2}$, so $N(d) \ | \ 8$. Since $3^x$ is odd, then its norm is odd and this implies that $N(d)=1$, i.e. $d=\pm 1$. So, $n-\sqrt{-2}$ and $n+\sqrt{-2}$ are coprime in $\mathbb{Z}[\sqrt{-2}]$. Since these two factors have the same norm and the only non trivial factorization (up to sign permutations) of $3$ with factors with the same norm is $3=(1-\sqrt{-2})(1+\sqrt{-2})$, then this forces
$$\begin{array}{rcl} n-\sqrt{-2}&=&(1-\sqrt{-2})^x \\ n+\sqrt{-2}&=&(1+\sqrt{-2})^x  \end{array} \qquad \textrm{or} \qquad \begin{array}{rcl} n-\sqrt{-2}&=&(1+\sqrt{-2})^x \\ n+\sqrt{-2}&=&(1-\sqrt{-2})^x  \end{array}$$
Considering the first equation, we get $\textrm{Im}(n-\sqrt{-2})=\textrm{Im}((1-\sqrt{-2})^x)$, i.e. $$-\sqrt{2}=\left[-{x \choose 1}+2{x \choose 3}-4{x \choose 5}+\ldots+(-1)^{\frac{x+1}{2}}2^{\frac{x-1}{2}}{x \choose x}\right]\sqrt{2},$$ i.e. $$-1=-{x \choose 1}+2{x \choose 3}-4{x \choose 5}+\ldots+(-1)^{\frac{x+1}{2}}2^{\frac{x-1}{2}}{x \choose x}.$$ Let $$\displaystyle f(x)=\sum_{k \textrm{ odd}}^x {x \choose k} (-1)^{\frac{k+1}{2}}2^{\frac{k-1}{2}}$$ be defined on the odd natural numbers. An easy check shows that $f(x) \leq f(x+2)$ for any odd $x$. Since $f(5)=11$, then $x \in \{1,3\}$. An easy check shows that $f(1)=f(3)=-1$ and we get $(x,y) \in \{(1,1),(3,1)\}$. If we consider the second system of equations, we get $\textrm{Im}(n-\sqrt{-2})=\textrm{Im}((1+\sqrt{-2})^x)$ and proceeding as before, we obtain no solutions.

(iii) If $y=0$, then $3^x-1=n^2$, i.e. $3^x=n^2+1$. Since $x>0$ and $n^2 \equiv 0,1 \pmod{3}$, we get no solutions in this case.

In conclusion, $(x,y) \in \{(1,1),(2,3),(3,1),(4,5)\}$.

Mathematical Reflections 2016, Issue 5 - Problem O385

Problem:
Let $f(x,y)=\frac{x^3-y^3}{6}+3xy+48$. Let $m$ and $n$ be odd integers such that $$|f(m,n)| \leq mn+37.$$ Evaluate $f(m,n)$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Let $m$ and $b$ be two odd integers such that $$-mn-37 \leq \dfrac{m^3-n^3}{6}+3mn+48 \leq mn+37.$$ We get the two inequalities
$$m^3-n^3+64+12mn \leq -2$$ and $$m^3-n^3+512+24mn \geq 2.$$ Using the identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ and setting $(a,b,c)=(m,-n,4)$ and $(a,b,c)=(m,-n,8)$, we obtain the two inequalities
$$
(m-n+4)(m^2+n^2+16+mn-4m+4n) \leq -2
$$
and
$$
(m-n+8)(m^2+n^2+64+mn-8m+8n) \geq 2
$$
Since $a^2+b^2+c^2-ab-bc-ca \geq 0$ with the equality if and only if $a=b=c$, then and as $m,n$ are odd integers, we obtain that $m^2+n^2+16+mn-4m+4n>0$ and $m^2+n^2+16+mn-4m+4n>0$. It follows that $m-n+4<0$ and $m-n+8>0$, i.e. $m-n+4 \leq -2$ and $m-n+8 \geq 2$, which gives $-6 \leq m-n \leq -6$, i.e. $m-n=-6$. Therefore, $$f(m,n)=-(m^2+mn+n^2)+3mn+48=-(m-n)^2+48=12.$$

Mathematical Reflections 2016, Issue 5 - Problem S389

Problem:
Let $n$ be a positive integer. Prove that for any integers $a_1, a_2, \ldots, a_{2n+1}$ there is a rearrangement $b_1, b_2, \ldots, b_{2n+1}$ such that $2^n\cdot n!$ divides $$(b_1-b_2)(b_3-b_4)\cdot \ldots \cdot (b_{2n-1}-b_{2n}).$$

Proposed by Cristinel Mortici, Valahia University, Targoviste, Romania

Solution:
Observe that we have $2n$ remainders modulo $2n$. By the pigeonhole principle there exist two integers among $a_1,a_2,\ldots,a_{2n+1}$, say $b_1$ and $b_2$, such that $(b_1-b_2)$ is divisible by $2n$. Now, discard $b_1$ and $b_2$. There remain $2n-1$ numbers among the ones in the given list. Since we have $2n-2$ remainders modulo $2n-2$, by the pigeonhole principle we have that there exist two integers, say $b_3$ and $b_4$, such that $(b_3-b_4)$ is divisible by $2n-2$. Proceeding in this way, we obtain a rearrangement $b_1, b_2, \ldots, b_{2n+1}$ of $a_1, a_2, \ldots, a_{2n+1}$ such that $$(b_1-b_2)(b_3-b_4)\cdot \ldots \cdot(b_{2n-1}-b_{2n})$$ is divisible by $2n\cdot (2n-2) \cdot (2n-4) \cdot \ldots \cdot 4 \cdot 2=2^n \cdot n!$.

Mathematical Reflections 2016, Issue 5 - Problem S385

Problem:
Let $a,b,c$ be positive real numbers. Prove that $$\dfrac{1}{a^3+8abc}+\dfrac{1}{b^3+8abc}+\dfrac{1}{c^3+8abc} \leq \dfrac{1}{3abc}.$$

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
The given inequality can be written as $$\dfrac{1}{9abc}-\dfrac{1}{a^3+8abc}+\dfrac{1}{9abc}-\dfrac{1}{b^3+8abc}+\dfrac{1}{9abc}-\dfrac{1}{c^3+8abc} \geq 0,$$ i.e. $$\dfrac{a^2-bc}{a^2+8bc}+\dfrac{b^2-ca}{b^2+8ca}+\dfrac{c^2-ab}{c^2+8ab} \geq 0.$$
Let $x=\dfrac{bc}{a^2}$, $y=\dfrac{ca}{b^2}$, $z=\dfrac{ab}{c^2}$. Then, we have $xyz=1$ and we want to prove that $$\dfrac{1-x}{1+8x}+\dfrac{1-y}{1+8y}+\dfrac{1-z}{1+8z} \geq 0.$$ Observe that $$\dfrac{1-x}{1+8x}+\dfrac{1-y}{1+8y}+\dfrac{1-z}{1+8z}=\dfrac{3(-64xyz+16(xy+yz+zx)+5(x+y+z)+1)}{(1+8x)(1+8y)(1+8z)}.$$ Since $xyz=1$, then by AM-GM Inequality we have $x+y+z \geq 3$ and $xy+yz+zx \geq 3$, so $$-64xyz+16(xy+yz+zx)+5(x+y+z)+1 \geq -63+16\cdot3+5\cdot3=0$$ and we get
$$\dfrac{1-x}{1+8x}+\dfrac{1-y}{1+8y}+\dfrac{1-z}{1+8z} \geq 0,$$ as we wanted to prove.

Mathematical Reflections 2016, Issue 5 - Problem J389

Problem:
Solve in real numbers the system of equations
$$(x^2-y+1)(y^2-x+1)=2[(x^2-y)^2+(y^2-x)^2]=4.$$

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Set $u=x^2-y$ and $v=y^2-x$. The given system can be rewritten as $$\begin{array}{rcl} u^2+v^2&=&2 \\ (u+1)(v+1)&=&4. \end{array} $$
Multiplying the second equation by $2$ and adding to the first equation, we get $$(u+v)^2+2(u+v)-8=0,$$ and solving for $u+v$, we get $u+v=-4$ or $u+v=2$. If $u+v=-4$, then $uv=4-u-v-1=7$. If $u+v=2$, then $uv=4-u-v-1=1$. We obtain the two systems of equations:
$$\begin{array}{rcl} u+v&=&-4 \\ uv&=&7, \end{array} \qquad \begin{array}{rcl} u+v&=&2 \\ uv&=&1. \end{array}$$
The first system gives no real solution. The second system gives $u=v=1$. We obtain the system of equations
$$\begin{array}{rcl} x^2-y&=&1 \\ y^2-x&=&1.\end{array}$$ Subtracting side by side the two equations, we get $(x^2-y^2)+(x-y)=0$, i.e. $$(x-y)(x+y+1)=0.$$ So, $x=y$ or $x=-1-y$. Substituting these values, we obtain $$(x,y) \in \left\{\left(\dfrac{1+\sqrt{5}}{2},\dfrac{1+\sqrt{5}}{2}\right),\left(\dfrac{1-\sqrt{5}}{2},\dfrac{1-\sqrt{5}}{2}\right),(0,-1),(-1,0),\right\}.$$

Mathematical Reflections 2016, Issue 5 - Problem J387

Problem:
Find all digits $a,b,c,x,y,z$ for which $\overline{abc},\overline{xyz}$ and $\overline{abcxyz}$ are all perfect squares (no leading zeros allowed).

Proposed by Adrian Andreescu, Dallas, Texas

Solution:
Let $\overline{abcxyz}=n^2$, where $n \in \mathbb{N}$. Since $\overline{abc}$ must be a perfect square, it's easy to see that $n \in \{317,348,349,380,412,443,475,506,538,570,601,633,696,728,759,791,854,886,949\}$. An easy check shows that the given conditions are satisfied if and only if $n \in \{380,475,570\}$. So, $$(a,b,c,x,y,z) \in \{(1,4,4,0,0,0),(2,2,5,6,2,5),(3,2,4,9,0,0)\}.$$

Mathematical Reflections 2016, Issue 5 - Problem J386

Problem:
Find all real solutions to the system of equations $$x+yzt=y+ztx=z+txy=t+xyz=2.$$

Proposed by Mohamad Kouroshi, Tehran, Iran

Solution:
Subtracting the second equation to the first equation, the third equation to the second equation, the fourth equation to the third equation and the first equation to the fourth equation, we obtain
$$\begin{array}{rcl} (x-y)(1-zt)&=&0 \\ (y-z)(1-tx)&=&0 \\ (z-t)(1-xy)&=&0 \\ (t-x)(1-yz)&=&0. \end{array}$$

We have four cases.

(i) $x-y=0$ and $z-t=0$, i.e. $x=y$ and $z=t$. Substituting these values into the second and the fourth equation, we get $(x-z)(1-zx)=0$. If $x=z$, then $x+x^3=2$, which gives $x=y=z=t=1$. If $zx=1$, then $y+t=2$, i.e. $x+z=2$, which gives $x=1,z=1$, so $x=y=z=t=1$.

(ii) $x-y=0$ and $1-xy=0$, i.e. $x=y$ and $xy=1$. We obtain $x=y=\pm 1$. If $x=y=1$, then $zt=1$ and $z+t=2$, which gives $z=t=1$. If $x=y=-1$, then $zt=-3$ and $z+t=2$, which gives $z=3,t=-1$ or $z=-1,t=3$.

(iii) $1-zt=0$ and $z-t=0$, i.e. $z=t$ and $zt=1$. We obtain $z=t=\pm 1$. If $z=t=1$, then $xy=1$ and $x+y=2$, which gives $x=y=1$. If $z=t=-1$, then $xy=-3$ and $x+y=2$, which gives $x=3,y=-1$ or $x=-1,y=3$.

(iv) $1-zt=0$ and $1-xy=0$, i.e. $zt=1$ and $xy=1$. We obtain $x+y=2$ and $z+t=2$, which gives $x=y=z=t=1$.

In conclusion, the real solutions to the given system of equations are $$(x,y,z,t) \in \{(1,1,1,1),(1,1,3,-1),(1,1,-1,3),(3,-1,1,1),(-1,3,1,1)\}.$$

Mathematical Reflections 2016, Issue 5 - Problem J385

Problem:
 If the equalities $$\begin{array}{rcl} 2(a+b)-6c-3(d+e)&=&6 \\ 3(a+b)-2c+6(d+e)&=&2 \\ 6(a+b)+3c-2(d+e)&=&-3 \end{array}$$ hold simultaneously, evaluate $a^2-b^2+c^2-d^2+e^2$.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Multiplying the second equation by $3$ and subtracting the first equation, we get $$a+b=-3(d+e).$$ Multiplying the third equation by $2$ and adding the first equation, we get $$2(a+b)=d+e.$$ It follows that $a+b=-6(a+b)$, which gives $a+b=0$ and $d+e=0$. So, $c=-1$ and
$$a^2-b^2+c^2-d^2+e^2=(a-b)(a+b)+c^2-(d-e)(d+e)=1.$$

Mathematical Reflections 2016, Issue 4 - Problem U381

Problem:
Find all positive integers $n$ such that
$$\sigma(n)+d(n)=n+100.$$

(We denoted by $\sigma(n)$ the sum of the divisors of $n$ and by $d(n)$ the number of divisors of $n$).

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Since $\sigma(n) \geq n+1$, then the sum of the proper divisors of $n$ is $99-d(n) \leq 97$. Denoting by $\omega(n)$ the number of distinct primes dividing $n$, observe that if $\omega(n) \geq 4$, then there exist four prime numbers $p<q<r<s$ that divide $n$. Since $p \geq 2$, $q \geq 3$, $r \geq 5$ and $s \geq 7$, then $qrs$ divides $n$ but $qrs \geq 3\cdot5\cdot7>97$, which contradicts the hypothesis. So, $\omega(n) \leq 3$. We have three cases.

(i) $\omega(n)=1$. Then, $n=p^k$, where $p$ is a prime number and $k \in \mathbb{N}^*$. If $k=1$, then $(1+p)+2=p+100$, contradiction. If $k \geq 2$, then $$1+p+\ldots+p^k+(k+1)=p^k+100,$$ i.e.
$$
p(1+p+\ldots+p^{k-2})+k=98. \qquad (1)
$$
If $k \geq 7$, then $$p(1+p+\ldots+p^{k-2}) \geq 2(1+2+\dots+2^5)=2\cdot63>97.$$ It follows that $k \in \{2,3,4,5,6\}$. An easy check in (1) shows that there are no solutions.

(ii) $\omega(n)=2$. Then, $n=p^{k_1}q^{k_2}$, where $p,q$ are prime numbers, $p<q$ and $k_1,k_2 \in \mathbb{N}^*$. Since $p \geq 2$ and $q \geq 3$, if $n=p^5q$, then $p^5+p^4q+p^3q \geq 32+16\cdot3+8\cdot3>97$, so $k_1 \leq 4$. If $n=pq^4$, then $q^4+q^3 \geq 81+27>97$, so $k_2 \leq 3$.  We have twelve cases.

(a) $k_1=1,k_2=1$. Then, $n=pq$ and $d(n)=4$. It follows that $p+q=95$. By parity, it must be $p=2$, but $q=93$ is not a prime. So, no solutions in this case.

(b) $k_1=2,k_2=1$. Then, $n=p^2q$ and $d(n)=6$. It follows that $p+p^2+q+pq=93$. By parity, it must be $p=2$ and we get $q=29$. So, $n=116$.

(c) $k_1=3, k_2=1$. Then, $n=p^3q$ and $d(n)=8$. It follows that $p+p^2+p^3+q+pq+p^2q=91$. By parity, it must be $p=2$ and we get $q=11$. So, $n=88$.

(d) $k_1=4, k_2=1$. Then, $n=p^4q$ and $d(n)=10$. It follows that $p+p^2+p^3+p^4+q+pq+p^2q+p^3q=89$. By parity, it must be $p=2$ and we get $q=59/11$, i.e. no solution.

(e) $k_1=1, k_2=2$. Then, $n=pq^2$ and $d(n)=6$. It follows that $p+pq+q+q^2=93$. By parity, it must be $p=2$ and we get $q(3+q)=91$, i.e. no solution.

(f) $k_1=2,k_2=2$. Then, $n=p^2q^2$ and $d(n)=9$. It follows that $p+p^2+pq+p^2q+pq^2+q+q^2=90$. By parity, it must be $p=2$ and we get $q(7+3q)=84$, i.e. no solution.

(g) $k_1=3, k_2=2$. Then, $n=p^3q^2$ and $d(n)=12$. But $$p^3q+p^2q^2+pq^2+p^2q \geq 8\cdot3+4\cdot9+2\cdot9+4\cdot3>99-d(n)=87,$$ so there are no solutions.

(h) $k_1=4, k_2=2$. Then, $n=p^4q^2$ and $d(n)=15$. But $$p^4+p^3q^2 \geq 16+8\cdot9>99-d(n)=84,$$ so there are no solutions.

(i) $k_1=1, k_2=3$. Then, $n=pq^3$ and $d(n)=8$. It follows that $p+pq+pq^2+q+q^2+q^3=91$. By parity, it must be $p=2$ and we get $q(3+3q+q^2)=89$, i.e. no solution.

(j) $k_1=2,k_2=3$. Then, $n=p^2q^3$ and $d(n)=12$. But $$q^3+pq^3+pq^2 \geq 27+2\cdot27+2\cdot9>99-d(n)=87,$$ so there are no solutions.

(k) $k_1=3, k_2=3$. Then, $n=p^3q^3$ and $d(n)=16$. But $$p^2q^3 \geq 4\cdot27>99-d(n)=83,$$ so there are no solutions.

(l) $k_1=4, k_2=3$. Then, $n=p^4q^3$ and $d(n)=20$. But $$p^2q^3 \geq 4\cdot27>99-d(n)=79,$$ so there are no solutions.

(iii) $\omega(n)=3$. Then, $n=p^{k_1}q^{k_2}r^{k_3}$, where $p,q,r$ are prime numbers, $p<q<r$ and $k_1,k_2,k_3 \in \mathbb{N}^*$. Since $p \geq 2$, $q \geq 3$ and $r \geq 5$, if $n=p^3qr$, then $p^3q+p^3r+p^2qr \geq 8\cdot3+8\cdot5+4\cdot3\cdot5>97$, so $k_1 \leq 2$. If $n=pq^2r$, then $pq^2+q^2r+pqr+r \geq 2\cdot9+9\cdot5+2\cdot3\cdot5>97$, so $k_2=1$. If $n=pqr^2$, then $qr^2+pr^2 \geq 3\cdot 25+2\cdot25>97$, so $k_3=1$.  We have two cases.

(a) $k_1=k_2=k_3=1$. Then, $n=pqr$ and $d(n)=8$. It follows that $p+q+r+pq+pr+qr=91$. By parity, it must be $p=2$, which gives $3q+3r+qr=89$, i.e. $(q+3)(r+3)=98$. There are no solutions in this case.

(b) $k_2=2, k_2=k_3=1$. Then, $n=p^2qr$ and $d(n)=12$. It follows that $p+p^2+q+pq+p^2q+r+pr+p^2r+qr+pqr=87$. By parity, it must be $p=2$, which gives $7q+7r+3qr=81$. If $q \geq 5$, then $r \geq 7$ and $7q+7r+3qr>81$. So, $q=3$ and $4r=15$, i.e. no solution.

In conclusion, $n \in \{88,116\}$.

Mathematical Reflections 2016, Issue 4 - Problem J382

Problem:
Find all triples $(x,y,z)$ of real numbers with $x,y,z>1$ satisfying
$$\left(\dfrac{x}{2}+\dfrac{1}{x}-1\right)\left(\dfrac{y}{2}+\dfrac{1}{y}-1\right)\left(\dfrac{z}{2}+\dfrac{1}{z}-1\right)=\left(1-\dfrac{x}{yz}\right)\left(1-\dfrac{y}{zx}\right)\left(1-\dfrac{z}{xy}\right).$$

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
By the AM-GM Inequality, we have
$$\dfrac{x}{2}+\dfrac{1}{x}-1=\left(\dfrac{x}{2}+\dfrac{1}{2x}\right)+\dfrac{1}{2x}-1 \geq \dfrac{1}{2x} > 0$$
$$\dfrac{y}{2}+\dfrac{1}{y}-1=\left(\dfrac{y}{2}+\dfrac{1}{2y}\right)+\dfrac{1}{2y}-1 \geq \dfrac{1}{2y} > 0$$
$$\dfrac{z}{2}+\dfrac{1}{z}-1=\left(\dfrac{z}{2}+\dfrac{1}{2z}\right)+\dfrac{1}{2z}-1 \geq \dfrac{1}{2z} > 0.$$
Thus, $$\left(1-\dfrac{x}{yz}\right)\left(1-\dfrac{y}{zx}\right)\left(1-\dfrac{z}{xy}\right)>0.$$
Assume that two among these three factors are negative. Without loss of generality, $1-\dfrac{x}{yz}<0$ and $1-\dfrac{y}{zx}<0$. Then, $\dfrac{x}{yz}>1$ and $\dfrac{y}{zx}>1$, which gives $\dfrac{1}{z^2}>1$, contradiction. Hence,
$$1-\dfrac{x}{yz}>0, \qquad 1-\dfrac{y}{zx}>0, \qquad 1-\dfrac{z}{xy}>0.$$
We have $$\begin{array}{lll} \left(\dfrac{x}{2}+\dfrac{1}{x}-1\right)^2-\left(1-\dfrac{y}{zx}\right)\left(1-\dfrac{z}{xy}\right)&=&\dfrac{1}{x^2}+\dfrac{2}{x}\left(\dfrac{x}{2}-1\right)+\left(\dfrac{x}{2}-1\right)^2-\left[1-\dfrac{yz}{x}\left(\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)+\dfrac{1}{x^2}\right] \\ &=&\left(\dfrac{x}{2}-1\right)^2+\dfrac{yz}{x}\left(\dfrac{1}{y}-\dfrac{1}{z}\right)^2 \geq 0. \end{array}$$
The equality holds if and only if $x=2$ and $y=z$. Likewise,
$$\left(\dfrac{y}{2}+\dfrac{1}{y}-1\right)^2 \geq \left(1-\dfrac{z}{xy}\right)\left(1-\dfrac{x}{yz}\right)$$
$$\left(\dfrac{z}{2}+\dfrac{1}{z}-1\right)^2 \geq \left(1-\dfrac{x}{yz}\right)\left(1-\dfrac{y}{zx}\right)$$
Multiplying these three inequalities, we get that the left-hand side of the given equation is greater or equal than the right-hand side and the equality holds if and only if $x=y=z=2$. So, $(x,y,z)=(2,2,2)$.