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Algebra Level 2

If x x and y y are positive numbers whose sum is 1, then which of the following inequalities must be true?

x 2 + y 2 ≥ x 3 + y 3 x^2 + y^2 \geq x^3 + y^3 x 3 + y 3 ≥ x 2 + y 2 x^3 + y^3 \geq x^2 + y^2

Pi Han Goh by Pi Han Goh , 30, Malaysia

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2 solutions

Maadhav Gupta
Oct 14, 2016

x + y = 1 x+y=1

x 2 + y 2 = ( x + y ) 2 − 2 x y = 1 − 2 x y x^2+y^2=(x+y)^2-2xy=1-2xy

x 3 + y 3 = ( x + y ) ( x − x y + y ) = ( 1 ) ( x 2 − x y + y 2 ) = ( x 2 + y 2 ) − x y = ( x + y ) 2 − 2 x y − x y = 1 − 3 x y x^3+y^3=(x+y)(x-xy+y)=(1)(x^2-xy+y^2)=(x^2+y^2)-xy=(x+y)^2-2xy-xy=1-3xy

Since 0 < x < 1 0<x<1 , 0 < y < 1 0<y<1 ,

2 x y < 3 x y 2xy<3xy

− 2 x y > − 3 x y -2xy>-3xy

1 − 2 x y > 1 − 3 x y 1-2xy>1-3xy

x 2 + y 2 > x 3 + y 3 x^2+y^2>x^3+y^3

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Well, here goes my first answer. If there's something wrong or any issue whatsoever then let me know. Also, I will begin learning to write in LaTex pretty soon.

Maadhav Gupta - 4 years, 8 months ago

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Nicely done! Keep it up!

Also, it should be a strict inequality instead as neither x nor y can be zero ("x and y are positive numbers").

Well, if a < b a<b and a ≠ b a\ne b is true, then a ≤ b a \leq b is also true.

Pi Han Goh - 4 years, 8 months ago

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Yes you are right.

Anandmay Patel - 4 years, 8 months ago

Oh I see. My bad. I always get confused in such inequalities and in the "is a subset of" part in Set Theory.

Maadhav Gupta - 4 years, 8 months ago

Since x + y = 1 and x an y are positive it must be that x,y <1. Now because of that x^3 < x ^2 and y^3 < y ^2. So it must hold true that x^3 + y^3 < x^2 + y^2. (Sorry because of mobile use less then or equal is not possible to type for me)

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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