Seems like Cauchy
The maximum value of f ( x , y , z ) = 5 x + 4 y + 4 z on the ellipsoid 1 0 x 2 + 2 0 y 2 + 1 5 z 2 = 1 can be expressed in the form n m , where m and n are coprime positive integers. Find m + n .
The answer is 161.
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2 solutions
As the problem name suggests, we can find the required answer very easily using Cauchy-Schwarz inequality. ( 1 0 x 2 + 2 0 y 2 + 1 5 z 2 ) ( 1 0 2 5 + 2 0 1 6 + 1 5 1 6 ) ≥ ( 5 x + 4 y + 4 z ) 2 . ( 1 ) ( 2 5 + 5 4 + 1 5 1 6 ) ≥ ( 5 x + 4 y + 4 z ) 2 ( 5 x + 4 y + 4 z ) ≤ 3 0 1 3 1 Hence the required answer is 1 3 1 + 3 0 = 1 6 1 .
By the way @Sean Ty Why is the problem under Number Theory section rather than Algebra??
@Calvin Lin Please change the topic..!
Similar solution as Prakhar Gupta 's
Using Cauchy-Schwarz inequality , we have:
( 1 0 5 ( 1 0 x ) + 2 0 4 ( 2 0 y ) + 1 5 4 ( 1 5 z ) ) 2 ( 5 x + 4 y + 4 z ) 2 ⟹ 5 x + 4 y + 4 z ≤ ( ( 1 0 5 ) 2 + ( 2 0 4 ) 2 + ( 1 5 4 ) 2 ) ( ( 1 0 x ) 2 + ( 2 0 y ) 2 + ( 1 5 z ) 2 ) ≤ ( 1 0 2 5 + 2 0 1 6 + 1 5 1 6 ) ( 1 0 x 2 + 2 0 y 2 + 1 5 z 2 ) ≤ ( 2 5 + 5 4 + 1 5 1 6 ) ( 1 ) ≤ 3 0 1 3 1 ≤ 3 0 1 3 1
⟹ m + n = 1 3 1 + 3 0 = 1 6 1