A4 Let x,y be positive reals satisfying the condition x3+y3≤x2+y2.Find the maximum value of xy.
Problem
Source: JBMO 2011 Shortlist A4
Tags: inequalities
15.05.2016 12:37
Murad.Aghazade wrote: A4 Let x,y be positive reals satisfying the condition x3+y3≤x2+y2.Find the maximum value of xy. Method 1: xy≤14(x+y)2≤(x3+y3)2(x2+y2)2≤(x2+y2)2(x2+y2)2=1Equality for x=y=1
15.05.2016 12:43
My solution by Chebishev we have x2+y2≥x3+y3≥(x+y)(x2+y2)2≥(xy)12⋅(x2+y2) A.M-G.M which means xy≤1
15.05.2016 13:02
Murad.Aghazade wrote: A4 Let x,y be positive reals satisfying the condition x3+y3≤x2+y2.Find the maximum value of xy. Method 2: Let y=kx, then we have: x≤k2+1k3+1, and: xy=kx2≤k(k2+1)2(k3+1)2=1−(k−1)2(k4+k3+k2+k+1)(k+1)2(k2−k+1)2≤1
15.05.2016 13:32
Murad.Aghazade wrote: A4 Let x,y be positive reals satisfying the condition x3+y3≤x2+y2.Find the maximum value of xy. Method 3: We can use the following inequality: ln(x)≤x−1≤x3−x2 which gives: ln(xy)=ln(x)+ln(y)≤x3−x2+y3−y2≤0Then we conclude that: xy≤1.
15.05.2016 16:35
Murad.Aghazade wrote: A4 Let x,y be positive reals satisfying the condition x3+y3≤x2+y2.Find the maximum value of xy. Method 4: if both x and y ≤1, the problem is obvious. Without loss of generality, we can assume that x≤y, then x≤1≤y. Let f(t)=t3−t2+x3−x2, then f is strictly increasing for t≥1 and we have f(y)≤0. If we assume that y>1x, than f(y)>f(1x)=(x−1)2(x4+x3+x2+x+1)x3≥0 Contradiction! then: y≤1x which means that xy≤1.
15.05.2016 16:52
Murad.Aghazade wrote: A4 Let x,y be positive reals satisfying the condition x3+y3≤x2+y2.Find the maximum value of xy. Method 5 Using Cauchy-Schawrtz, we can prove that: x3+y3≤x2+y2≤x+y≤2Then: xy≤x2+y22≤1. Equality for x=y=1.
15.05.2016 20:07
Murad.Aghazade wrote: A4 Let x,y be positive reals satisfying the condition x3+y3≤x2+y2.Find the maximum value of xy. Method 6: We can use Power Mean: 2√x2+y22≤3√x3+y32≤3√x2+y22Then we conclude that x2+y2≤2 and by AM-GM xy≤x2+y22≤1Equality for x=y=1.
23.09.2023 11:46
bel.jad5 wrote: Murad.Aghazade wrote: A4 Let x,y be positive reals satisfying the condition x3+y3≤x2+y2.Find the maximum value of xy. Method 5 Using Cauchy-Schawrtz, we can prove that: x3+y3≤x2+y2≤x+y≤2Then: xy≤x2+y22≤1. Equality for x=y=1. hey can u pls elaborate this solution