Minimising can be Difficult!
For real numbers x , y , z , subject to x + y + z > 6 and
x 2 + y 2 + z 2 − x y z = ( x + y ) ( y + z ) ( z + x ) − 8 ( x y + y z + z x ) ,
what is the minimum value of x y + y z + z x ?
The answer is 24.
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2 solutions
Very nice use of the identity ( x + y ) ( y + z ) ( z + x ) + x y z = ( x + y + z ) ( x y + y z + z x ) :)
W h y n o t s i m p l y A 2 − 2 4 A + 1 4 4 = 0 ⟹ A − 6 A 2 = 2 4 = B , ∵ a − 6 > 0 ? ? How to find values of x, y, z? How can we say that it is the minimum? P l e a s e h e l p .
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I'm not sure if this answers your question, but the condition that you found ( a − 6 > 0 ) is simply a condition for the expression to be valid; it says nothing about the minimum value.
The quadratic inequality that I used is to show that B ≥ 2 4 . But why would that mean that 2 4 is the minimum? It's because this tells us that no value of B can be less than 2 4 , and since it is achievable (as there exist x , y , z that work), it must be the minimum.
In fact the values of x , y , z are not unique; I just gave them as an example so that A = 1 2 , B = 2 4 . Hope this helps :)
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Thanks. But part of my question was how do you get values of x, y, z knowing A=12, B=24. Thank you.
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@Niranjan Khanderia – I just set x = 0 then solved for y + z = 1 2 , y z = 2 4 to give variables that satisfy the conditions. There are infinitely many different solutions; I just gave an example where x = 0 to show that the minimum was possible.
With x y z + (x + y)(y + z)(z + x) = (x + y + z)(x y + y z + z x), {Identity by Daniel Liu }
x y + y z + z x = (x + y + z)^2/ (x + y + z - 6)
Basically, S^2/ (S - 6) >= 24 with S > 6 and (x y + y z + z x) is positive.
With z = 0, I actually simulated and found at least (0.00, 2.45, 9.80) for 24.01,
x y = (x + 24/ x)^2/ (x + 24/ x - 6) = 24
x^4 - 24 x^3 + 192 x^2 - 576 x + 576 = 0 {Over expanded.}
(x^2 - 12 x + 24)^2 = 0
[(x - 6)^2 - 12]^2 = 0
[x - 6 + 2 Sqrt (3)]^2 [x - 6 - Sqrt (3)]^2 = 0
Hence, with
x = 2 (3 - Sqrt (3)) = 2.535898384862245412945107
y = 2 (3 + Sqrt (3)) = 9.464101615137754587054893
z = 0
Substitute into original equations, satisfied. Therefore minimum of x y + y z + z x = 24.
Let x + y + z = A , x y + y z + z x = B for brevity. Using the identity ( x + y ) ( y + z ) ( z + x ) + x y z = ( x + y + z ) ( x y + y z + z x ) we can rearrange to give
A 2 = A B − 6 B ⟹ B = A − 6 A 2
We also know that
A 2 − 2 4 A + 1 4 4 ≥ ( A − 1 2 ) 2 ≥ 0 ⟹ A 2 ≥ 2 4 ( A − 6 )
But we are given that A > 6 so we can divide through by A − 6 to give
A − 6 A 2 ≥ 2 4 ⟹ B ≥ 2 4
Also note that this minimum is achieved when x = 0 , y = 2 ( 3 + 3 ) , z = 2 ( 3 − 3 ) so the answer is 2 4 .