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

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