Stronger than Classical Inequalities!
Let a , b , and c be real numbers such that 0 ≤ a , b , c ≤ 2 1 and a + b + c = 1 . Find the maximum value of
a 3 + b 3 + c 3 + 4 a b c
The answer is 0.28125.
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2 solutions
Nice solution. After substituting c = 1 - a - b, the inequalities became 0<=a,b<=1/2 and a+b>=1/2. I had a hunch that I only needed to check the three boundaries (actually only two because two were identical w.r.t the problem) of the 2-D inequalities. I thought about trying Lagrange multipliers, but I guess I was just lazy. I was more interested in finding an algebra solution. I never found one. But I wonder if it can be easily justified that all you need to do is check the boundary.
f ( a , b , c ) = a 3 + b 3 + c 3 + 4 ∗ a ∗ b ∗ c . f ( 1 / 2 , 0 , 1 / 2 ) = . 2 5 . f ( 1 / 3 , 1 / 3 , 1 / 3 ) = . 2 5 9 2 5 9 2 5 9 . . . . . f ( 1 / 2 , 1 / 3 , 1 / 6 ) = . 2 7 7 7 7 7 . . . f ( 1 / 2 , 1 / 4 , 1 / 4 ) = . 2 8 1 2 A l l p o s s i b l e p a t t e r n s h a s b e e n i n v e s t i g a t e d . S o f o r t h e g i v e n c o n d i t i o n s , m a x i m u m p o s s i b l e v a l u e s a r e g i v e n t o a , b , c , t h a t r e t u r n s f m a x = . 2 8 1 2 5 . M a x v a l u e i s g i v e n t o a = 1 / 2 . M a x p o s s i b l e v a l u e s t o b a n d c c a n o n l y b e 1 / 4 s o t h a t n o n e g e t s s m a l l v a l u e .