2006 Romanian NMO Problems/Grade 9/Problem 1

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

Find the maximal value of

$\left( x^3+1 \right) \left( y^3 + 1\right)$,

where $x,y \in \mathbb R$, $x+y=1$.

Dan Schwarz

Solution

If y is negative, then $(x^3+1)(y^3+1)$ is also negative, so we want $0\leq y\leq 1$.

$y=\dfrac{1}{a}$

where $1 \leq a$. Let's see what happens when a gets large:

$(x^3+1)(y^3+1)=\left(\dfrac{(a-1)^3}{a^3}+1\right)\left(\dfrac{1}{a^3}+1\right)=\dfrac{(a^3+1)((a-1)^3+a)}{a^6}$

$=\dfrac{a^6-3a^5+4a^4-3a^2+4a-1}{a^6}=1-\dfrac{3a^5-4a^4+3a^2-4a+1}{a^6}$

As a gets large, the fraction gets small, therefore maximizing $(x^3+1)(y^3+1)$. But when a gets small(up to 2), the fraction gets bigger, and therefore lessens $(x^3+1)(y^3+1)$.

Therefore, the maximum value of $(x^3+1)(y^3+1)$ is when x=1 and y=0, which is 2.