For x, y in C,

Consider simultaneous equations of x and y

x + y = u

xy = v.

Let the solutions be x = α and y = β.

Thus

α+β = u

αβ = v.

A quadratic equation whose solutions are α and β is

t² -ut+v =0.

By α, β in C, u and v are also in C.

Here, considering α, β in R, the solutions of t² -ut+v =0 must be real solutions.

Hence u²-4v≧0 holds.

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By

x + y = u

xy = v, we have

x²-ux+v=0.

By x in R,

u² - 4v ≧ 0