From a math paper, how do they get this inequality? It's probably simple but I'm not sure why I'm not seeing it. The inequality is from a paper ("Multiplicative-noise Can Suppress Chaotic Oscillation in Lotka-Volterra Type Competitive Model"):
$$\begin{align*}
\sum_{i=1}^4 \rho_i (x_i-1)(1-\sum_{j=1}^4 \alpha_{ij}x_j) &\leq\begin{split}
&\mathbin{\hphantom{+}}x_1(2\rho_1 + \rho_2\alpha_{21}+ \rho_3\alpha_{31}+\rho_4\alpha_{41}) \\&+x_2(\rho_1\alpha_{12} + 2\rho_2+ \rho_3\alpha_{32}+\rho_4\alpha_{42}) \\&+ x_3(\rho_1\alpha_{13} + \rho_2\alpha_{23}+ 2\rho_3+\rho_4\alpha_{43}) \\&+ x_4(\rho_1\alpha_{14} + \rho_2\alpha_{24} + \rho_3\alpha_{34}+2\rho_4)
\end{split}
\end{align*}$$
The paper has what I believe to be a typo with an extra $2$ in the last line so what is above is what the inequality should be. Basically, each $x_i(t)$ is the size of a species at time (a continuous variable between $0$ and $\infty$), and the other terms represent the growth rate ($\rho$) and strength of competition between them $\alpha$. They say this inequality is due to the positivity of  $x_i(t)$. I know $\rho_i > 0$, and I believe each $\alpha$ should be positive as well. I'm not understanding how they were able to get this inequality although I believe it to be true, and it would be very useful to understand this for a paper I'm trying to write. I'm guessing it has something to do with the range where $x_i-1$ and $1-\sum_{j=1}^4 \alpha_{ij}x_j$ are both negative so their product becomes positive, but I'm just not seeing it. If anyone is able to explain where this is coming from, it would be very much appreciated :)
 A: We move everything on the R.H.S., obtain a linear expression in the $\rho$-variables.
Let us look specifically at the part / at the coefficient of $\rho_1$, denote below with $E_1$. 
(From the given data it is well possible to pass to the limit $\rho_1\to\infty$, and in the same time $\rho_2,\rho_3,\rho_4\to0$. We thus need in the given generality $E_1\ge 0$, and corresponding inequalities for the other parts / coefficients in $\rho_2,\rho_3,\rho_4$, respectively.)
So $E_1$ is:
$$
\begin{aligned}
E_1
&=
(2x_1
+a_{12}x_2
+a_{13}x_3
+a_{14}x_4)
+
(x_1-1)
\Big(
a_{11}x_1 +
a_{12}x_2 +
a_{13}x_3 +
a_{14}x_4
-1
\Big)
\\
&=
(2x_1
\color{blue}{
+a_{12}x_2
+a_{13}x_3
+a_{14}x_4})
+
x_1
\Big(
a_{11}x_1 +
a_{12}x_2 +
a_{13}x_3 +
a_{14}x_4
-1
\Big)
\\
&\qquad\qquad-
\Big(
a_{11}x_1 +
\color{blue}{
a_{12}x_2 +
a_{13}x_3 +
a_{14}x_4}
-1
\Big)
\\
&=
a_{11}(x_1^2-x_1)+x_1
\
+
a_{12}x_1x_2 +
a_{13}x_1x_3 +
a_{14}x_1x_4
\
+1\ .
\end{aligned}
$$
Now if the "size" is an integer, the above is $>0$ since $x_1^2-x_1\ge 0$ on $0,1,2,3,4,\dots$, but if the "size" is really a continuous variable between $0$ and $\infty$, then some choices of the letters may make the expression negative. 
(Take for instance $x_1=\frac 12$, $a_{11}\to\infty$, and $x_2,x_3,x_4\to0$.)
(There must be some hidden conditions somewhere in the continuous case.)
