Prove $m^2 \ge 4kx$ for Real Solutions of $x^3+mx+k=0$

[anemone]

Prove that any real solution of $x^3+mx+k=0$ satisfies the inequality $m^2 \ge 4kx$.

 

[Euge]

Spoiler

If $x$ is a real root of the cubic equation, then the quadratic equation $a^2 + ma + kx = 0$ has a real root, namely $a = x^2$. Thus, the discriminant of the quadratic is nonnegative. This implies $m^2 \ge 4kx$.

 

[anemone]

Spoiler

If $x$ is a real root of the cubic equation, then the quadratic equation $a^2 + ma + kx = 0$ has a real root, namely $a = x^2$. Thus, the discriminant of the quadratic is nonnegative. This implies $m^2 \ge 4kx$.

Awesome, Euge! Thanks for participating and thanks for your great solution too! :)