How to check if $x^TCx\geq0$? I have the next 3x3 block matrix C, where each block is a square matrix.
$$
C = \begin{bmatrix}
0 & A & B \\
0 & A+K_1 & B \\
0 & A & B+K_2
\end{bmatrix},
$$
where $K_i$ is also a squared diagonal matrix (I have freedom choosing their elements). I want to be sure that $x^TCx \geq 0$ choosing the appropriated $K_i$.
EDIT
The answer is no
 A: $$
C = \begin{bmatrix}
0 & A & B \\
0 & A+K_1 & B \\
0 & A & B+K_2
\end{bmatrix},
$$
Let $A$ be a $n\times n$ matrix.
Subtract $1$st row from $(n+1)$ th row then from By a matrix multiplying the matrix $P_1$ from left of $C$(Let $C_1=PC$). 
Now we will do the same thing but now we will subtact $2$ nd row from $n+2$ th row by multiplying with the matrix $P_2$.In this way we will go on i.e we will subtract the $i$ th row from $(n+i)$th row using the matrix $P_i$.So we have $P_nP_{n-1}\dots P_{1}C=C_n=\begin{bmatrix}
0 & A & B \\
0 & K_1 &0 \\
0 & A & B+K_2
\end{bmatrix}$
Now again,
Subtract $1$st row from $(2n+1)$ th row then from By a matrix multiplying the matrix $Q_1$ from left of $C$(Let $C_{n+1}=PC_1$). 
Now we will do the same thing but now we will subtact $2$ nd row from $2n+2$ th row by multiplying with the matrix $Q_2$.In this way we will go on i.e we will subtract the $i$ th row from $(2n+i)$th row using the matrix $Q_i$.So we have $Q_nQ_{n-1}\dots Q_{1}P_nP_{n-1}\dots P_{1}C=C_{2n}=\begin{bmatrix}
0 & A & B \\
0 & K_1 &0 \\
0 & 0 & K_2
\end{bmatrix}$
Now Let $D=Q_nQ_{n-1}\dots Q_{1}P_nP_{n-1}\dots P_{1}$
Here $C_{2n}$ is triangular ,so $C_{2n}D^t=\begin{bmatrix}
0 & A & B \\
0 & K_1 &0\\
0 & 0 & K_2
\end{bmatrix}$ is also triangular(Use the fact that multiplying $C_{i-1}$ by $P_i^t$ from right which will subtract the ist column from the $n+i$ column of $C_{i-1}$.And so for $Q_i$)
So we have $DCD^t=\begin{bmatrix}
0 & A & B \\
0 & K_1 &0\\
0 & 0 & K_2
\end{bmatrix}$
A: Unless $A=B=0$, this is an impossible mission.
Let $x^\top =(u^\top ,v^\top ,w^\top )$, where the partitioning of $x$ conforms to the partitioning of $C$. If $x^\top Cx\ge0$ for all $x$, then in particular, this is true for $w=0$, i.e. $x^\top Cx=u^\top Av + v^\top (A+K_1)v\ge0$. However, if $A\ne0$, then for every $v\notin\ker(A)$, we can always find a vector $u$ (e.g. $u=-kAv$ for a large positive $k$) such that $u^\top Av$ and in turn $x^\top Cx$ are as negative as possible.
Put it another way, $x^\top Cx\ge0$ for all $x$ if and only if $P=C+C^\top $ is positive semidefinite. For a positive semidefinite matrix $P$, if it has a zero on its diagonal, the row and column containing this zero diagonal entry must also be zero, otherwise $P$ will contain a $2\times2$ submatrix that has negative determinant. Now the $(1,1)$-th subblock of $C+C^\top $ is zero, but the $(1,2)$-th and $(1,3)$-th subblocks are $A$ and $B$. So, $C+C^\top $ cannot be positive semidefinite when $A\ne0$ or $B\ne0$.
