Solve in positive real numbers the system of equations: <mtable rowspacing="3pt" columnspacing=

Raul Walker 2022-07-01 Answered
Solve in positive real numbers the system of equations:
( x 3 + y 3 ) ( y 3 + z 3 ) ( z 3 + x 3 ) = 8 , x 2 x + y + y 2 y + z + z 2 z + x = 3 2 .
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Answers (1)

verzaadtwr
Answered 2022-07-02 Author has 17 answers
Note that
y 2 x + y + z 2 y + z + x 2 z + x = ( x 2 x + y + y x ) + ( y 2 y + z + z y ) + ( z 2 z + x + x z ) = x 2 x + y + y 2 y + z + z 2 z + x = 3 2 .
Therefore,
x 2 + y 2 x + y + y 2 + z 2 y + z + z 2 + x 2 z + x = 3 .
By the AM-GM Inequality, we see that
(*) ( x 2 + y 2 x + y ) ( y 2 + z 2 y + z ) ( z 2 + x 2 z + x ) 1 .
We claim that, for a , b > 0,
(#) a 3 + b 3 2 ( a 2 + b 2 a + b ) 3 ,
and the inequality becomes an equality if and only if a = b. To justify the claim, let p ( t ) R [ t ] be the polynomial
p ( t ) := 2 ( t 2 + 1 ) 3 ( t + 1 ) 3 ( t 3 + 1 ) .
Since p ( 1 ) = 16 16 = 0, we know that ( t 1 ) is a factor of p ( t ). Now,
p ( t ) = 12 t ( t 2 + 1 ) 2 3 ( t + 1 ) 2 ( t 3 + 1 ) 3 t 2 ( t + 1 ) 3 ,
which satisfies p ( 1 ) = 48 24 24 = 0 again. That is, ( t 1 ) 2 is a factor of p ( t ). We proceed further:
p ( t ) = 48 t 2 ( t 2 + 1 ) + 12 ( t 2 + 1 ) 2 6 ( t + 1 ) ( t 3 + 1 ) 18 t 2 ( t + 1 ) 2 6 t ( t + 1 ) 3 .
We have again that p ( 1 ) = 96 + 48 24 72 48 = 0, and so ( t 1 ) 3 is a factor of p ( t ). Now,
p ( t ) = 144 t ( t 2 + 1 ) + 96 t 3 6 ( t 3 + 1 ) 54 t 2 ( t + 1 ) 54 t ( t + 1 ) 2 6 ( t + 1 ) 3 ,
so p ( 1 ) = 288 + 96 12 108 216 48 = 0, whence ( t 1 ) 4 is a factor of p ( t ). Because p is a monic polynomial of degree 6, we must have
p ( t ) = ( t 1 ) 4 ( t 2 + α t + β )
for some α , β R. With p ( 0 ) = 1, we get β = 1. As p ( 1 ) = 16, we conclude that 1 α + β = 1, so that α = 1, as well. Consequently,
p ( t ) = ( t 1 ) 4 ( t 2 + t + 1 ) ,
which is a nonnegative polynomial (i.e., p ( R ) R 0 ), and the only real root of p ( t ) is t = 1.
Now, (#) is equivalent to
2 ( a 2 + b 2 ) 3 ( a + b ) 3 ( a 3 + b 3 ) = b 6 p ( a b ) = ( a b ) 4 ( a 2 + a b + b 2 ) 0 ,
which is an equality iff a = b. Hence, the claim is established, but then we conclude that
( x 3 + y 3 2 ) ( y 3 + z 3 2 ) ( z 3 + x 3 2 ) 1 ,
using (#) in (*). Therefore,
( x 3 + y 3 ) ( y 3 + z 3 ) ( z 3 + x 3 ) 8 .
However, the problem statement demands that the inequality above is an equality. That is, x = y = z must hold. Ergo, the only positive real solution to this system of equations is
( x , y , z ) = ( 1 , 1 , 1 ) .

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