Intermediate problems (1)
I would be posting a series of intermediate-leveled problems for enjoyment.
Find all solutions to the following equations in reals:
x 1 + y 2 + z 3 = 1
x y z = 1 6 2
∣ x − ∣ x ∣ ∣ + ∣ y − ∣ y ∣ ∣ + ∣ z − ∣ z ∣ ∣ = 0
If there are n solutions are in the form ( x , y , z ) = ( x i , y i , z i ) , i = 1 , 2 , 3 , … , n , find i = 1 ∑ n x i
This problem is part of the set
The answer is 3.
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2 solutions
y z + 2 x z + 3 x y = 1 6 2 x y < 5 3 x ( 2 z + 3 y ) < 1 6 2 x < = 1 6 Try on with x=2,3,6,18 With only some solution x=3, y=6, z=9 satisfy the problem i = 1 . . n ∑ ( x i ) = 3
x, y, z are not necessarily integers. Although it works in this case it kight not work if x, y, z are non integral
From the equation with the absolute values, we can infer that x, y, and z are all positive. However, the first equation is rather unfriendly; we seek a more elegant representation.
This inspires one to substitue: z = 3 u y = 2 v
Thus, x y z = 1 6 2 implies that x ( 2 v ) ( 3 u ) = 6 ( x v u ) = 1 6 2 and x v u = 2 7
By the AM - GM -HM inequality, we have that x 1 + v 1 + u 1 3 < = ( x v u ) 3 1 = 2 7 3 1 = 3 Therefore, since the left hand side of the above inequality is three, and the equality case of AM - GM - HM occurs only when x = v = u, we know that x = v = u = 3. So, x = 3