Proving the difference of two matrices is PSD Claim: For $x\in \mathbb{R}^n$, we have  $\operatorname{Diag}(x) - xx^T \succeq 0$ if and only if $x_i \geq 0 \ \forall i\in [n]$ and $\sum_{i} x_i \leq 1$.
Where $\operatorname{Diag}$ denotes the diagonal matrix. How can I prove this? 
I tried principal minor test (induction etc.), doesn't work well. Also, tried to decompose to the eigenvalues, not a nice spectral decomposition. My only hope is using the traditional definition: for every $a \in \mathbb{R}^n_{+}$, we have the following equivalence   $a^TXa\geq0 \iff X\succeq0$. 
When I follow the traditional definition, I come to:
$\sum_{i=1}^n \sum_{j=1}^n a_ia_j(-x_ix_j) + \sum_{i=1}^na_i^2x_i \geq 0$
but stuck again..
 A: $(\Longrightarrow)$ Let $e_i$ be the $i^{th}$ standard basis vector. If $\def\D{\operatorname{Diag}(x)}\D-xx\def\t{^\top}\t\succeq0,$ then
$$
e_i\t\D e_i-e_i\t x x\t e_i=x_i-x_i^2=x_i(1-x_i)\ge 0\implies 0\le x_i\le 1.
$$
Furthermore, letting $\def\1{{\bf 1}}\1$ be the all ones vector, then
$$
\1\t \D \1-\1\t xx\t  \1=(\sum_i x_i)-(\sum_ix_i)^2\ge 0\implies 0\le \sum_i x_i\le 1.
$$
$(\Longleftarrow)$ For any $a\in\mathbb R^n$, let $\def\a{\overline{a}}\a\in \mathbb R^n$ be the constant vector whose entries are all equal to $\sum_i x_ia_i=a\t x$. Then
$$
\begin{align}
0&\le (a-\a)\t\D (a-\a)
\\
&= a\t \D a-a\t\D \a -\a \t \D a\t+\a\t \D \a
\\
&\stackrel{\star}= a\t \D a-\;\;\;\;\;(a\t x)^2\;\;\; -\;\;\;\;(a\t x)^2\;\;\;\;\;+(a\t x)^2\cdot(\textstyle\sum_i x_i)
\\
&\le a\t \D a-\;\;\;\;\;(a\t x)^2\;\;\; -\;\;\;\;(a\t x)^2\;\;\;\;\;+(a\t x)^2
\\
&= a\t \D a-(a\t x)^2
\\
&= a\t \D a-a\t xx\t a
\end{align}
$$ 
The magic of the proof is hidden away in $\stackrel{\star}=$; take some time to convince yourself that step is correct. 
A: You can use Schur complements, schur complement wikipedia. An example of the 2 dimensional case:
$$
\begin{align}
Diag(x) - xx^T &\succeq0 \\
\begin{bmatrix}
x_0-x_0^2 & -x_0x_1 \\
-x_0x_1 & x_1-x_1^2 \\
\end{bmatrix} =  
\begin{bmatrix}
A & B \\
B^T & C \\
\end{bmatrix}
& = X \succeq0
\end{align}
$$ 
So from Schur's complement $A \succeq0$ then $X \succeq0 \iff C-B^TA^{-1}B \succeq0$
Solving $C-B^TA^{-1}B$ we get:
$$\frac{(x_1x_0)(1-x_0-x_1)}{(x_0-x_0^2)}$$
From the above equation, $x_i \geq 0 \ \forall i\in [n]$ and $\sum_{i} x_i \leq 1$, we can see that: $(x_1x_0)$ and $(1-x_0-x_1)$ are greater than or equal to zero and that $(x_0-x_0^2)$ is greater than 0 but less than 1. This means that $C-B^TA^{-1}B \succeq0$ which means that $X \succeq0$.
Can you generalize this for the nxn case?
