Find the minimum of f

Algebra Level 3

Let a , b , c , d a,b,c,d be real numbers greater or equal to -1, and whose sum is 2.

Minimize a 3 + b 3 + c 3 + d 3 a^3 + b^3 + c^3 + d^3 .

0 1 2 \frac12 3 2 \frac32 Doesn't exist

Othman Slassi by Othman Slassi , 34, Morocco

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1 solution

Othman Slassi
Jan 16, 2018

( ∀ x ∈ [ − 1 , + ∞ [ ) : ( x − 1 2 ) 2 ( x + 1 ) = x 3 − 3 4 x + 1 4 ≥ 0 (\forall x \in [-1, +\infty[): (x-\frac{1}{2})^{2}(x+1)=x^3-\frac{3}{4}x+\frac{1}{4}\ge 0

So if ( a , b , c , d ) ∈ I (a,b,c,d)\in I we sum up the inequalities for x = a , b , c , d x=a,b,c,d and we get f ( a , b , c , d ) ≥ 1 2 f(a,b,c,d)\ge \frac{1}{2} with equality occurs when a = b = c = d = 1 2 a=b=c=d=\frac{1}{2}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

( ∀ x ∈ [ − 1 , + ∞ [ ) : ( x − 1 2 ) 2 ( x + 1 ) = x 3 − 3 4 x + 1 4 ≥ 0 (\forall x \in [-1, +\infty[): (x-\frac{1}{2})^{2}(x+1)=x^3-\frac{3}{4}x+\frac{1}{4}\ge 0

How did you come up with this statement?

Pi Han Goh - 3 years, 4 months ago

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(x-1/2)^2 is positive, (x+1) is positive too, the product is a positive real number.

Othman Slassi - 3 years, 4 months ago

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