It isn't as complicated as it looks
( 1 − x ) ( 1 − y ) ( 1 − z ) 1 + ( 1 + x ) ( 1 + y ) ( 1 + z ) 1 For − 1 < x , y , z < 1 , let t denote the minimum value of the above expression and u be the sum of x , y , z when the equality holds, find ( 3 t ) u + 4 .
The answer is 5.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
Can you explain more why ( 1 − x 2 ) ( 1 − y 2 ) ( 1 − z 2 ) ≤ 1 ?
Log in to reply
Cause x 2 ≥ 0 ⇒ − x 2 ≤ 0 ⇔ 1 − x 2 ≤ 1 , y and z are similar
Since x , y , z ∈ ( − 1 ; 1 ) , the expression is positive and finite. Because, the symmetry with respect to the variables, the solution requires x , y and z to be equal at the minimum. Because the symmetry with respect to signs, the solution must be with each variable being 0 . t = 2 . Therefore, u = 0 . Therefore, the solution is 5.
This is a close parallel solution to P C's solution.
Since x , y , z ∈ ( − 1 ; 1 ) , both fraction above are positive
Applying AM-GM we get ( 1 − x ) ( 1 − y ) ( 1 − z ) 1 + ( 1 + x ) ( 1 + y ) ( 1 + z ) 1 ≥ ( 1 − x 2 ) ( 1 − y 2 ) ( 1 − z 2 ) 2 Notice that ( 1 − x 2 ) ( 1 − y 2 ) ( 1 − z 2 ) ≤ 1
So t = 2 and the equality holds when x = y = z = 0 ⇒ u = 0 ( 3 t ) u + 4 = 1 + 4 = 5