The answer to both your questions are yes. Let me start with the first question, which more straightforward.
A first remark: since $A$ is positive-definite, $\lambda_i(A+B) \geq \lambda_i(B)$ for $i=1,2$.
(to check this, use the formulas $\lambda_1(X) = \max_{\xi} \langle X\xi,\xi\rangle$ and $\lambda_2(X) = \min_{\xi} \langle X\xi,\xi\rangle$ where the min and max run over all unit vectors $\xi$. This formulas hold whenever $X$ is a symmetric $2 \times 2$ matrix.)
Your question is therefore whether $Tr(\sqrt{A+B})\leq Tr(\sqrt A)+ Tr(\sqrt B)$ for any symmetric positive definite matrices $A$ and $B$, or equivalently $Tr( \sqrt{X X^{T} + Y Y^{T} } ) \leq Tr(\sqrt{X X^{T} })+Tr(\sqrt{Y Y^{T} })$ for any matrices $X$ and $Y$, where $X^{T}$ denotes the transpose of $X$, or hermitian tranpose if you work with complex matrices. This inequality is true in any dimension (not just 2), and it is just the triangle inequality for the Schatten 1-norm given by $\|X\|_1 = Tr(\sqrt{X X^{T} })$. The expression $Tr(\sqrt{X X^{T} + Y Y^{T} })$ is indeed the 1-norm of the matrix $\begin{pmatrix}X&Y \\\\ 0&0\end{pmatrix}$.
EDIT: It seems from the comments that my answer to your second question was far from clear. Let me try to explain differently the proof I had in mind.
For 6 real numbers $\alpha_1 \geq \alpha_2$, $\beta_1\geq \beta_2$ and $\gamma_1 \geq \gamma_2$, denote by $f(\alpha_1,\alpha_2,\beta_1 , \beta_2,\gamma_1,\gamma_2)$ the quantity $\sqrt{|\alpha_1|} + \sqrt{|\alpha_2|} - |\sqrt{\max(|\gamma_1|,|\gamma_2|)} - \sqrt{\max(|\beta_1|,|\beta_2|)}| - |\sqrt{\min(|\gamma_1|,|\gamma_2|)} - \sqrt{\min(|\beta_1|,|\beta_2|)}|$.
You are asking whether $f \geq 0$ provided that $\alpha,\beta,\gamma$ are the ordered eigenvalues of respectively $A,B,A+B$ for symmetric $2 \times 2$ matrices $A$ and $B$. The answer is yes, and I am sketching a proof. Denote by $D$ the possible values for $(\alpha_1,\alpha_2,\beta_1 , \beta_2,\gamma_1,\gamma_2)$.
$D$ is exactly described by Horn's inequalities. These inequalities are
$$\alpha_1 \geq \alpha_2 \ \ , \ \ \beta_1\geq \beta_2,$$
$$\gamma_1 + \gamma_2= \alpha_1 + \alpha_2+\beta_1+\beta_2,$$
$$\alpha_2+\beta_2 \leq\gamma_2 \leq \min(\alpha_1+\beta_2,\alpha_2+\beta_1).$$
In particular, $D$ is a convex subset of dimension $5$ of $\mathbb R^6$, and one easily checks that its boundary corresponds to the case when $A$ and $B$ commute. Since the inequality is true when $A$ and $B$ commute (this is eay to check, see the other answer), your question reduces to whether $\inf_D f = \inf_{\partial D} f$. This transforms your eigenvalue question to a purely calculus question.
Notice now that $\beta,\gamma$ and $\alpha_1+\alpha_2$ being fixed, $f(\alpha,\beta,\gamma)$ decreases as $\min(|\alpha_1|,|\alpha_2|)$ decreases. Moreover, if you started with $\alpha,\beta,\gamma$ in the interior of $D$, you stay in $D$ if you make $\min(|\alpha_1|,|\alpha_2|)$ decrease, until you reach the boundary of $D$, or $\min(|\alpha_1|,|\alpha_2|)=0$. You are therefore left to prove that $f(\alpha,\beta,\gamma) \geq \inf_{\partial D} f$ if $(\alpha,\beta,\gamma) \in D$ with $\min(|\alpha_1|,|\alpha_2|)=0$.
In the same way, fixing $\alpha,\beta$ and $\gamma_1+\gamma_2$, you reduce the question to proving that $f(\alpha,\beta,\gamma) \geq \inf_{\partial D} f$ if $(\alpha,\beta,\gamma) \in D$ with $\min(|\alpha_1|,|\alpha_2|)=0$ and $\min(|\gamma_1|,|\gamma_2|)=0$.
Last, fixing $\alpha, \gamma$ and $\beta_1+\beta_2$ with $\min(|\alpha_1|,|\alpha_2|)=0$ and $\min(|\gamma_1|,|\gamma_2|)=0$, you see that $f(\alpha,\beta,\gamma)$ decreases as $\min(|\beta_1|,|\beta_2|)$ increases, until you reach the boundary of $D$. This proves that $\inf_D f = \inf_{\partial D} f$.
Best Answer
Consider the following matrix
$a b^T + b a^T$ where $a=\Delta$, $b = P(\Phi+\Phi^T)\Delta$. Both $a$ and $b$ are N by 1 matrix.
Consider a vector $v = a + x b$ where $x$ is a real number. we have
$ (a b^T + b a^T )v = [(a b^T + b a^T )a] + x[(a b^T + b a^T )b] $
$ =[a (b^T a) + b (a^T a)] + x[a (b^T b) + b (a^T b)] $
$ = [ (b^T a) + x (b^T b) ] a + [ (a^T a)+ x(a^T b)]b $
$ =(b^T a +x b^T b) a + (a^T a+ x b^T a) b$
Note that $(b^T a) = (a^T b) $ , $a^T a$, and $(b^T b)$ are all real numbers. The idea to construct $v$ this way is because the column space of $X$ is spanned from $a,b$, so the eigenvector must be the linear combination of $a$ and $b$.
Let
$ (b^T a +x b^T b) = \lambda $
and
$ (a^T a+ x b^T a) = \lambda x $
We have $(a^T a+ x b^T a) = x (b^T a +x b^T b) $
$ x^2(b^T b) - a^T a = 0 $
So we have the two roots of $x= \pm \sqrt{\frac{a^T a}{b^T b}}$
So $\lambda = b^T a \pm \sqrt{a^T a \times b^T b}$ and by cauchy schwarz
$\lambda_{max} = b^T a + \sqrt{a^T a \times b^T b} > 0$
$\lambda_{min} = b^T a - \sqrt{a^T a \times b^T b} < 0$
So the conclusion is that the two nonzero eigenvalues has the opposite sign.