math question ( calculus)?

Question

Which number(s)when increased by its reciprocal and three times the square of its reciprocal will always generate a value never less than 13/4 ?(please explain how u get the answer)

Answer 1

x is a number, where it's inverse times 3 is 3/x
so your looking for (3/x)+(3/x)^2>13/4
now lets just take the integral of that from o to a so that the integral = 13/4
so (-3/a^2)+(-9/a^3)=13/4
solve for a. by 0=13a^3-12a-36.
Find those zeros.

Answer 2

x + 1/x + 3/x^2 ≥ 13/4
4x^3 + 4x + 12 ≥ 13x^2
4x^3 - 13x^2 + 4x + 12 ≥ 0
x^3 - (13/4)x^2 + x + 3 ≥ 0
(x + 3/4)(x - 2)^2 ≥ 0
x ≥ - 3/4