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The inequality is given by $$x^{H}(\Phi(x))^{-1}x-x^{H}\dfrac{aa^H}{a^H\Phi(x)a}x\ge0,\text{ for any }x\ne0,$$ where $\Phi(x)$ is positive definite and is a function of $x$, $a$ can be any nonzero vector.
The equality holds only when $x=\Phi(x)a$.
I know that the condition $\Phi(x)^{-1}-\dfrac{aa^H}{a^H\Phi(x)a}\succeq0 $ is sufficient but not necessary. However, I checked it numerically and found the condition always held.
Thank you very much!!!
$\Phi(x)$ is positive definite, so it has a unique positive definite square root $P$ and the inequality in question can be rewritten as $$ \|P^{-1}x\|^2 \|Pa\|^2 - |\langle x,a\rangle|^2 \ge 0.\tag{1} $$ Why is this true? As $P$ is Hermitian, $\langle x,a\rangle=\langle P^{-1}x,\,Pa\rangle$. Therefore $(1)$ is merely Cauchy-Schwarz inequality and the condition that equality holds is $P^{-1}x=Pa$, i.e. $x=\Phi(x)a$.
