Prove that for all positive real numbers x and y, 1/x + 1/y + xy ≥ 3, with equality if and only if x = y = 1?

Question

Madhukar · Accepted Answer

For the three numbers, 1/x, 1/y and xy,
Arithmetic Mean ≥ Geometric Mean
=> (1/3) (1/x + 1/y + xy) ≥ [(1/x) * (1/y) * xy]^(1/3)
=> 1/x + 1/y + xy ≥ 3.

Edit:
Equality is obvious for x = y = 1

Anonymous · Answer

There are an limitless variety of recommendations to this subject. eg: x=a million/3, y=a million/3 eg: x=0, y=-a million/3 eg: x=-a million/3, y=0 Substituting in any value for x yields a cubic - which continuously has a minimum of one actual answer (for y).

Prove that for all positive real numbers x and y, 1/x + 1/y + xy ≥ 3, with equality if and only if x = y = 1?

2 Answers