Question

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Answer 1

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Answer 2

Răspuns:

(x,y,z) ∈ {(3,2,4);(6,4,8);(9,6,12);(12,8,16)}

Explicație pas cu pas:

$\frac{x}{3} = \frac{y}{2} = \frac{z}{4} = k, \ k \not = 0 \\ \implies x = 3k; \: y = 2k; \: z = 4k$

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} > \frac{1}{4} \iff \frac{1}{3k} + \frac{1}{2k} + \frac{1}{4k} > \frac{1}{4} \\ \frac{4 + 6 + 3}{12k} > \frac{1}{4} \iff \frac{13}{12k} > \frac{1}{4} \\ k < \frac{13 \cdot 4}{12} \iff k < \frac{13}{3} = 4 \frac{1}{3} \\ \implies k \in \{1,2,3,4\}$

$k = 1 \implies x = 3; \ y = 2; \ z = 4$

$k = 2 \implies x = 6; \ y = 4; \ z = 8$

$k = 3 \implies x = 9; \ y = 6; \ z = 12$

$k = 4 \implies x = 12; \ y = 8; \ z = 16$

$\bf (x,y,z) \in \{(3,2,4);(6,4,8);(9,6,12);(12,8,16) \}$