Olympiad Problem 5
f ( x ) = x 3 ( x 3 + 1 ) ( x 3 + 2 ) ( x 3 + 3 )
Find the minimum value of the function above.
The answer is -1.
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2 solutions
It is simpler to let x 6 + 3 x 3 = t . Then we have t ( t + 2 ) = t 2 + 2 t ≥ − 1 , since t 2 + 2 t + 1 = ( t + 1 ) 2 ≥ 0 .
nice solution
why is my approach incorrect? by am>gm y=x^3 (y+y+1+y+2+y+3)/4 >= (f(x))^1/4
(y+6/4) >(f(x))^1/4 f(x) < (y+6/4)^1/4
f(x) =0 for y=x^3 = -6/4
Done the same way.
y = x 3 − 2 3 f ( x ) = ( y − 2 3 ) ( y − 2 1 ) ( y + 2 1 ) ( y − 2 3 ) = ( y 2 − 4 9 ) ( y 2 − 4 1 ) = y 4 − 2 5 y 2 + 1 6 9 f ( x ) = ( y 2 − 4 5 ) 2 − 1 for real y, f ( x ) m i n = 0 − 1 = − 1 the question must say real x.
f ( x ) = ( x 3 ) ( x 3 + 3 ) ( x 3 + 1 i s ( x 3 + 2 ) f ( x ) = ( x 6 + 3 x 3 ) ( x 6 + 3 x 3 + 2 )
Let x 6 + 3 x 3 + 1 = t
f ( x ) = t 2 − 1
So to minimize f ( x ) we need to minimize ∣ t ∣
And ∣ a ∣ > 0 or = 0 for any real a
t = x 6 + 3 x 3 + 1
Observe its discriminant D = 3 2 − 4 × 1 × 1 = 5 > 0
So minimum ∣ t ∣ could be 0 at x 3 = 2 − 3 ± 5 which is possible.
Implies f ( x ) m i n . = − 1