Given that
⎩ ⎪ ⎨ ⎪ ⎧ tan − 1 a + tan − 1 b + tan − 1 c = 2 π sin − 1 x + sin − 1 y + sin − 1 z = 2 π
If 8 x y z = 1 , then find the minimum value of x a 2 + y b 2 + z c 2 .
This is an original problem and belongs to the set My Creations .
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tan − 1 a + tan − 1 b + tan − 1 c = 2 π ⟹ a b + b c + c a = 1
sin − 1 x + sin − 1 y + sin − 1 z = 2 π ⟹ x 2 + y 2 + z 2 = 1 − 2 x y z
From Titu's Lemma , we have x a 2 + y b 2 + z c 2 ≥ x + y + z ( a + b + c ) 2 .
So for x a 2 + y b 2 + z c 2 to be minimum, ( a + b + c ) should be minimum and ( x + y + z ) has to be maximum.
Consider { a , b , c } and { b , c , a }
Applying Cauchy-Schwarz Inequality we get,
( a + b + c ) 2 ≥ ( a b + b c + c a ) 2
⟹ ( a + b + c ) ≥ 1
Therefore, minimum value of ( a + b + c ) is 1 .
Consider { x , y , z } and { 1 , 1 , 1 }
Applying Cauchy-Schwarz Inequality we get,
3 ( x 2 + y 2 + z 2 ) ≥ ( x + y + z ) 2
3 ( 1 − 2 x y z ) ≥ ( x + y + z ) 2
But given that x y z = 8 1
So we obtain ( x + y + z ) 2 ≤ 3 ( 1 − 8 2 )
⟹ ( x + y + z ) 2 ≤ 3 ( 1 − 4 1 )
⟹ ( x + y + z ) ≤ 2 3
Therefore, maximum value of ( x + y + z ) is 2 3
Finally substituting the two obtained values in the required expression, we get
x a 2 + y b 2 + z c 2 ≥ 2 3 ( 1 ) 2
⟹ x a 2 + y b 2 + z c 2 ≥ 3 2
Hence, the minimum value of x a 2 + y b 2 + z c 2 is 3 2 .