2006 Romanian NMO Problems/Grade 9/Problem 1

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.

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2006 Romanian NMO Problems/Grade 9/Problem 1

Problem

Solution

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