a b c + ( 1 − a ) ( 1 − b ) ( 1 − c ) .
For a , b , c ∈ [ 0 , 1 ] , find the maximum value of the expression above.
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.
After applying AM-GM , why will the maximum be 1? I mean you showed that equality occurs when α = β = γ = 0 but why would it be the maximum?
Log in to reply
Can you complete your solution so that it seems more justified to me...Because I don't see any reason why it has to be the maximum and not any other value instead of 1
Log in to reply
Frankly, I cannot, else I would have added in the solution.
Log in to reply
@Chew-Seong Cheong – Usually when we have established an inequality such as f ( x , y , z ) ≥ g ( x , y . z ) , we just need to check if there are values of ( x , y , z ) such that f ( x , y , z ) = g ( x , y , z ) to show the maximum or minimum. Note that if you put x = y = z = 0 or 2 π . Both sides equal and in both cases the value is 1 which means the answer is unique.
Problem Loading...
Note Loading...
Set Loading...
Since a , b , c ∈ [ 0 , 1 ] , we can let a = sin 2 α , b = sin 2 β and c = sin 2 γ . Then
X = a b c + ( 1 − a ) ( 1 − b ) ( 1 − c ) = sin α sin β sin γ + cos α cos β cos γ
By AM-GM inequality,
X = sin α sin β sin γ + cos α cos β cos γ ≤ 3 sin 3 α + sin 3 β + sin 3 γ + 3 cos 3 α + cos 3 β + cos 3 γ
And equality occurs when α = β = γ = 0 or 2 π , and the maximum values of X is 1 .