crazyfootballer Says
I was not able to understand the solution. Got lost in between.
Ok...please take a pen and paper and solve the polynomials,
I've marked in bold, the end result of polynomials subtraction.
We have basically combined the terms from the result in such a manner to prove that
B^(x,y,z) - A(x,y,z) is always positive in both the scenarios when x>= 2z and when x<2z.
Let x > y > z > 0 and x >= 2z, then
B(x,y,z) = x+y-z,
if (B(x,y,z))^2 > A(x,y,z) => (B(x,y,z))^2 - A(x,y,z) > 0, solving we get
2y(3x-y) + 2x(3y-2z) which is > 0 as
3x-y > 0 and 3y-2z>0 Let x > y > z > 0 and x < 2z, then
B(x,y,z) = y+z+(2z-x) = 3z +y -x
if (B(x,y,z))^2 > A(x,y,z) => (B(x,y,z))^2 - A(x,y,z) > 0, solving we get
(3z +y -x)^2 - (x - 3y + 2z)(x-y) > 0 => 9z^2 - 2y^2 +8yz +2xy -8zx > 0
=>
z(9z-8x+8y) + 2x(x-y) > 0, which we can see always is as
x>y, and
9z-8x+8y > 0(why?)