Easiest way to show positive semi-definite equivalence For an $x \in \mathbb{R}^n$, and $n$-by-$n$ identity matrix $I_n$, we are given that
$$ \begin{pmatrix} I_n & x \\ x^T & 1 \end{pmatrix} \succeq 0.$$
What is the easiest way to show that
$$ \begin{pmatrix} 1 & x^T \\ x & I_n \end{pmatrix} \succeq 0$$
holds?
 A: This is due to the identity
$$ \underbrace{\begin{pmatrix} 0_{1,n} & 1 \\ I_n & 0_{n,1} \end{pmatrix}}_{J^T} \underbrace{\begin{pmatrix} I_n & x \\ x^T & 1 \end{pmatrix}}_{A} \underbrace{\begin{pmatrix} 0_{n,1} & I_n \\ 1 & 0_{1,n} \end{pmatrix}}_{J}  = \underbrace{\begin{pmatrix} 1 & x^T \\ x & I_n \end{pmatrix}}_{B} $$
(notation $0_{m,n}$ is for a zero block with $m$ lines and $n$ columns).
Indeed, $J$ being a permutation matrix, it is an orthogonal matrix, with $J^T=J^{-1}$. We can conclude that $A$ and $B$ are similar, thus have the same spectrum (Similar matrices have the same eigenvalues with the same geometric multiplicity) with positive eigenvalues, thus are both semi-definite positive. 
Besides, $A$ being symmetric, one can conclude from $J^TAJ=B$ that $B$ is symmetric as well.

Appendix : There is a pending question : is there a criteria on $x$ for positive semi-definiteness of $A$. ? The answer is yes :

$A$ is semi-definite positive iff $\|x\| \leq 1$.

This is due, as we are going to see it, to an analysis of the rather particular spectrum of $A$. Let us obtain it explicitly.
First of all, let us establish that $A$ (which is a $(n+1) \times (n+1)$ matrix) has eigenvalue $1$ with order of multiplicity at least $n-1$. 
Consider hyperplane $x^{\perp}$ of $\mathbb{R}^n$ defined as the set of vectors $y$ that are orthogonal to $x$. Let $(y_1,y_2,\cdots y_{n-1})$ be a basis of $x^{\perp}$ ; then, 
$$\underbrace{\begin{pmatrix} I_n & x \\ x^T & 1 \end{pmatrix}}_{A}\underbrace{\begin{pmatrix} y_k\\ 0 \end{pmatrix}}_{V_k}=1\underbrace{\begin{pmatrix} y_k\\ 0 \end{pmatrix}}_{V_k} \ \ \ \text{for} \ \ k=1,2, \cdots (n-1),$$
proving that $V_k$ is an eigenvector associated with eigenvalue $1$.
Due to the fact that trace$(A)=n+1$, the two remaining eigenvalues are of the form $\alpha$ and $\beta:=2-\alpha$. We can assume, WLOG that $\alpha \leq 1 \leq \beta$.
Besides, using the so-called Schur determinant identity (Eigenvalues of a Block Matrix from Schur Determinant Identity) for the computation of the determinant of a $2 \times 2$ block matrix, we obtain :
$$\det(A)=1-x^Tx$$
As the determinant is also the product of eigenvalues, we get the following identity :
$$\det(A)=1-\|x\|^2=\alpha(2-\alpha)\tag{1}$$
Thus, one can compute explicitly the two remaining eigenvalues by solving quadratic equation (1), with the following explicit solutions (if we assume that $\alpha$ is the smallest eigenvalue)
$$\alpha=1 - \|x\| \ \ \ \implies \ \ \ \beta:=2-\alpha=1 + \|x\|\tag{2}$$
As the criteria for a symmetric matrix do be semi-definite positive is that must have all eigenvalues $\geq 0$, this criteria becomes $\alpha \geq 0$, i.e., $\|x\| \leq 0$. $\square$
Remark : eigenvalues $\alpha$ and $\beta$ can be associated with eigenvectors $\begin{pmatrix} x\\ -\|x\| \end{pmatrix}$ and $\begin{pmatrix} x\\ \|x\| \end{pmatrix}$ resp. 
Let us take an example in the case $n=4$ ; let $m=1/n$ ; consider matrix :
$$A:=\left(\begin{array}{rrrr|r}
 1 & & & & m \\
 & 1 & & & m \\
& & 1 & & m\\
 & & & 1 & m\\
\hline
m & m & m & m & 1 \\
\end{array}\right)$$
One can check, using (2), that the spectrum of $A$ is 
$$(\tfrac12, 1 , 1, 1, 1, \tfrac32).$$
Just now, I "googled" with keywords "bordered identity matrix" : I found in (Eigenvalues of a certain bordered identity matrix) a somewhat similar computation that I did in the Appendix.
