A.Maths Inequalities

Let X and B be the roots of the quadratic equation x^2-[m+(3/m)]x+2=0 where m is a positive constant

(a) Using the fact that (a-b)^2>=0 for all real numbers a and b, show that m^2+(9/m^2)>=6.Hence show that [m+(3/m)]^2>=12
X+B=[m+(3/m)]------------1
XB=2-------------------2
(X-B)^2=(X+B)^2-4XB>=0
m^2+2m(3/m)+(9/m^2)-4(2)>=0
m^2+(9/m^2)>=2
m^2+6+(9/m^2)-6>=2
[m+(3/m)]^2>=8
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(a)
m^2+(9/m^2)-6
=(m^4-6m^2+9)/m^2
=(m-3)^2/m^2
>=0
Since (m-3)^2>=0, m^2>=0
So m^2+(9/m^2)>=6
Then
[m+(3/m)]^2-12
=m^2+6+(9/m^2)-12
=m^2+(9/m^2)-6
>=0 by previous result
So [m+(3/m)]^2>=12