16. Given the equation about $x$, $x^{2}+4 x+3 k-1=0$, the sum of the squares of its two real roots is not less than the product of these two roots, and for the inverse proportion function $y=\frac{1+5 k}{x}$, the graph's two branches in their respective quadrants show that the ordinate $y$ decreases as the abscissa $x$ increases. Find the integer values of $k$ that satisfy the above conditions.
16. From the problem, the equation
$$
x^{2}+4 x+3 k-1=0
$$

has real roots, so $\Delta=4^{2}-4(3 k-1) \geqslant 0$.
Solving this, we get $k \leqslant \frac{5}{3}$.
Let $x_{1} 、 x_{2}$ be the roots of (1), by the relationship between roots and coefficients, we know
$$
x_{1}+x_{2}=-4, x_{1} x_{2}=3 k-1 \text {. }
$$

Since $x_{1}^{2}+x_{2}^{2} \geqslant x_{1} x_{2}$, we have
$\left(x_{1}+x_{2}\right)^{2}-3 x_{1} x_{2} \geqslant 0$, which means $(-4)^{2}-3(3 k-1) \geqslant 0$.
Solving this, we get $k \leqslant \frac{19}{9}$.
Also, $y=\frac{1+5 k}{x}$, when $x>0$ or $x<0$, we have $1+5 k>0$, which means
$$
k>-\frac{1}{5} \text {. }
$$

Combining (2), (3), and (4), we get $-\frac{1}{5}<k \leqslant \frac{5}{3}$.
Therefore, the integer values of $k$ that satisfy the conditions are 0, 1.