An inequality on the real part of a square root

Question

I have the following inequality:

$\Re(k+z) \geq \Re \sqrt{(k+z)^2-4z}$

where $k$ is real and $z$ complex. Under what conditions on $k$ and $z$ is this inequality true? I suspect that it is true for $k>1$ and $\Re z > 0$, but don't know to prove it.

Answer 1

Consider the polynomial equation

$w^2 + (k + z) w + z = 0$.

Since it has the solutions $w = \frac{1}{2}(-k-z \pm \sqrt{(k+z)^2 - 4 z)})$, the inequality in the question is true if the solutions $w$ have negative real part. Applying the Routh-Hurwitz theorem for polynomials with complex coefficients, we can establish this by showing that the leading principal minors $P_{11}$ and $P_{33}$ of the matrix

$P = \pmatrix{k+\Re(z) & -\Im (z) & 0 \\ 1 & -\Im (z) & - \Re (z) \\ 0 & k + \Re(z) & -\Im(z)}$

are positive. Indeed, we have

$P_{11} = k + \Re(z)$

and

$P_{33} = \det P = \Im(z)^2 (k + \Re (z)) + \Re(z)(k + \Re(z))^2 - \Im(z)^2$

which are both positive for $k > 1$ and $\Re(z) > 0$.