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suppose a and b are real numbers such that the roots of the cubic equation ax^3-x^2+bx-1=0 are all positive real numbers. Prove that: (i) 0<3ab\le1 and (ii) b\ge \sqrt{3}
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2 Answers
21By Viete's formulae we have (x,y,z being roots)
x+y+z=1/a
xy+yz+zx = b/a
xyz=1/a
Now,
(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy+yz+zx)
1/a^2 = x^2 + y^2 +z^2 + 2b/a
Thus,
(1-2ab)/a^2 = x^2 + y^2 + z^2
Now by Q.M. >= A.M
we have
(1-2ab)/3a^2 >= 1/9a^2
3-6ab >= 1
3ab=<1
And as a and b are all positive 3ab > 0 is easily proved.
now (xy + yz + zx)/xyz = b
1/x + 1/y + 1/z = b
Again x+y+z / xyz = 1/xy + 1/yz + 1/zx = 1
Thus we have,
1/x^2 + 1/y^2 + 1/z^2 = b^2 - 2
Now by Cauchy Schwartz inequality
(1/x^2 + 1/y^2 + 1/z^2)(1^2 + 1^2 + 1^2)>= (1/x.1 + 1/y.1 + 1/z.1)^2
3(b^2 - 2)>= b^2
2b^2>=6
b>= root(3)
Proof completed