Suppose that $x_1,\ldots,x_4$ are nonnegative, I'm trying to prove that $$ \frac{x_2+x_1x_3+x_2x_4}{1+x_1^2+x_2^2+x_3^2+x_4^2}\leq\frac{9}{10} $$ which is equivalent to $$ 9(1+x_1^2+x_2^2+x_3^2+x_4^2)-10(x_2+x_1x_3+x_2x_4)\geq 0.\tag{i} $$ Unable to simplify the LHS of (i) (mostly because of not having "enough" of $x^2$), I split the problem into 2 cases: $x_1x_3\geq x_2x_4$ and $x_2x_4\geq x_1x_3$. I can prove (i) for the former but not the latter. Can you please help me? I hope I might see a handling of the case $x_2x_4\geq x_1x_3$ or a more elegant approach to the problem.
More context: you probably already recognize the LHS of the original inequality as the autocorrelation of order $2$ for the $MA(4)$ process $Y_t=(1+x_1L+x_2L^2+x_3L^3+x_4L^4)u_t$ where $u_t$ is white noise and $L$ as usual denotes the lag operator.
We have: $$ 5(x_1^2+x_3^2)\geq 10 x_1 x_3, $$ so we just need to prove: $$ 9(1+x_2^2+x_4^2)\geq 10 x_2(1+x_4). \tag{1}$$ We can regard both terms as functions of $x_2$ for a fixed $x_4$, so we just need to prove that $$ 9(1+q^2)+9x^2 \geq 10(1+q) x \tag{2}$$ for any $x\geq 0$ (and any $q\geq 0$). This is a simple convexity inequality: the derivative of the LHS is $10(1+q)$ for $x=\frac{5}{9}(1+q)$, and the LHS is a convex function, hence we just need to check that $(2)$ holds for $x=\frac{5}{9}(1+q)$. The resulting inequality is equivalent to: $$ 81(1+q^2)\geq 25(1+q)^2 \tag{3}$$ that is very easy to check.