By (here stood a wrong reasoning with a correct result) $ k\le 9$. For $ k = 1,2,3,4,5,6,8,9$ we easily find examples $ (9,9,9),(4,4,8),(3,3,3)(2,2,4),(1,4,5),(1,2,3),(1,1,2),(1,1,1)$ respectively. For $ k = 7$, there is no solution, but the reasoning was wrong so check freemind's proof below.

Among all positive integer triples satisfying $ \dfrac{(x + y + z)^2}{xyz} = k$ take the one with smallest $ \max\{x,y,z\}$. But Peter correctly got $ k\le9$. Consider the equation as a quadratic trinomial in $ x$. Then $ x^2 + x(2y + 2z - kyz) + (y + z)^2 = 0$. This has also the positive integer root $ kyz - 2y - 2z = \dfrac{(y + z)^2}x$. According to our choice of $ (x,y,z)$ we have $ z\le\dfrac{(x + y)^2}z$, hence $ z\le x + y$. Take now $ f(z) = \dfrac{(x + y + z)^2}{xyz} = z + \dfrac{(y + x)^2}{z} + 2(x + y)$. $ f'(z) = 1 - \dfrac{(x + y)^2}{z^2}$, thus the function attains its minimum at $ x + y$, hence it's decreasing on $ [y,x + y]$, from where $ k = f(z)\le\dfrac{(x + 2y)^2}{xy^2}$. Now, using $ x\le y$, we get $ k\le\dfrac{9y^2}{xy^2}\le9$, hence $ k\le9$. For $ k = 1,2,3,4,5,6,8,9$ we may use the examples Peter provided above. Let's handle $ k = 7$. Note that in the last inequality if $ 2\le x$ then we get $ k\le\dfrac92$. This means $ x = 1$. Now $ z\le 1 + y$ means $ z\in\{y,y + 1\}$. Both cases are handled easily leading to no solution.

freemind wrote: Take now $ f(z) = \dfrac{(x + y + z)^2}{xyz} = z + \dfrac{(y + x)^2}{z}+ 2(x + y)$. $ f'(z) = 1 - \dfrac{(x + y)^2}{z^2}$, thus the function attains its minimum at $ x + y$, hence it's decreasing on $ [y,x + y]$, from where $ k = f(z)\le\dfrac{(x + 2y)^2}{xy^2}$. Now, using $ x\le y$, we get $ k\le\dfrac{9y^2}{xy^2}\le9$, hence $ k\le9$. $ 9yz-(x+y+z)^2=3(x+y-z)(y+z-2x)+(2z-x-y)(4y+z-5x)\geq 0$, hence $ k=\frac{(x + y + z)^2}{xyz}\leq\frac{9}{x}\leq 9$, and $ x\leq\frac{9}{k}$.

Let $ S_k$ be the set of all triples of positive integers $ (a,b,c)$, $ a \leq b \leq c$, such that $ k = \frac {(a + b + c)^2}{abc}$. Suppose that $ S_k$ is non-empty. Take any $ (a,b,c) \in S_k$ such that $ a + b + c$ is minimized. Without loss of generality, let $ a \geq b \geq c$. Note that $ a^2 - a(kbc - 2b - 2c) + (b + c)^2 = 0$. The equation $ x^2 - x(kbc - 2b - 2c) + (b + c)^2$ has two (not necessarily distinct) solutions, one of which is $ a$; let the other be $ r$. By Vieta's formulas, we have that $ ar = (b + c)^2$ and $ a + r = kbc - 2b - 2c$. Note that $ r \geq a$, or else $ (r,b,c) \in S_k$ and $ r + b + c < a + b + c$, contradicting the minimality of $ a + b + c$. $ a \geq b \geq \frac {b + c}{2} \geq c$, so $ (b + c)^2 = ar \geq r\left(\frac {b + c}{2}\right)$, giving $ r \leq 2(b + c)$. In addition, $ a^2 \leq ar = (b + c)^2$, so $ a \leq b + c$. Therefore, $ kbc - 2b - 2c = a + r \leq (b + c) + 2(b + c) = 3(b + c)$. Rearrange this to get that $ (kb - 5)(kc - 5) \leq 25$. It follows that at least one of $ |kb - 5|, |kc - 5| \leq 5$ (since the other can be zero), giving $ 0 \leq kb \leq 10$. Since $ k \neq 0$ and $ bk \geq k$, $ 1 \leq k \leq 10$. If $ k = 10$, then $ 10b - 5$ and $ 10c - 5$ are both nonzero. In addition, $ 10b - 5, 10c - 5 \geq 5$, yielding $ (10b - 5)(10c - 5) \geq 25$, with equality holding exactly when $ b = c = 1$. Since $ (10b - 5)(10c - 5) \leq 25$, we must have $ b = c = 1$. Hence, $ 10 = \frac {(a + 1 + 1)^2}{a} = \frac {a^2 + 2a + 1}{a} = a + 2 + \frac {1}{a}$. Hence, $ a | 1$, so $ a = 1$; clearly, the right-hand side is not 1, so $ k \neq 10$. If $ k = 7$, then $ 7b - 5$ and $ 7c - 5$ are both nonzero. If $ c \geq 2$, then $ 7b - 5 \geq 7c - 5 \geq 9$, so $ 25 \geq (7b - 5)(7c - 5) \geq 81$, a contradiction. Hence, $ c = 1$, so $ 7b - 5 \leq \frac {25}{7\cdot 1 - 5} < 13$. Therefore, $ b < \frac {18}{7} \leq 3$, so $ b = 1$ or $ b = 2$. If $ b = 1$, then $ 7 = \frac {(a + 1 + 1)^2}{a} = a + 2 + \frac {1}{a}$, implying that $ a = 1$, resulting in the left hand side being 5, which is a contradiction. If $ b = 2$, then $ 7 = \frac {(a + 3)^2}{2a}$, so $ a^2 + 6a + 9 = 14a$, that is, $ a^2 - 8a + 9 = 0$. But the discriminant of this quadratic is $ 64 - 4 \cdot 9 = 28$, which is not a perfect square, so $ a$ is not an integer. Hence, $ k \neq 10$. There are solutions for all $ k$ between 1 and 9 inclusive. Indeed, $ (9, 9, 9)$ yields $ k = 1$, $ (4, 4, 8)$ yields $ k = 2$, $ (3, 3, 3)$ yields $ k = 3$, $ (2, 2, 4)$ yields $ k = 4$, $ (1, 4, 5)$ yields $ k = 5$, $ (1, 2, 3)$ yields $ k = 6$, $ (1, 1, 2)$ yields $ k = 8$, and $ (1, 1, 1)$ yields $ k = 9$. Hence, the set of all integers that can be taken on by $ \frac {(a + b + c)^2}{abc}$ is $ \{1, 2, 3, 4, 5, 6, 8, 9\}$.

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by vieta jumping $ k\leq 9$

$ k= \{1, 2, 3, 4, 5, 6, 8, 9\}$