2. Find all positive real numbers $t$ such that for any non-negative real number $x$, the inequality

$$
\frac{t}{x+2}+\frac{x}{t(x+1)} \leqq 1
$$

holds.
2. Thanks to the conditions $x \geq 0$ and $t>0$, we can equivalently transform the given inequality into a quadratic inequality in terms of $k x$:

$$
\begin{aligned}
\frac{t}{x+2}+\frac{x}{t(x+1)} & \leq 1, \\
t^{2}(x+1)+x(x+2) & \leq t(x+1)(x+2), \\
x^{2}+\left(t^{2}+2\right) x+t^{2} & \leq t x^{2}+3 t x+2 t, \\
0 & \leq (t-1) x^{2}+\left(3 t-t^{2}-2\right) x+2 t-t^{2}, \\
0 & \leq (t-1) x^{2}+(2-t)(t-1) x+t(2-t) .
\end{aligned}
$$

By substituting $x=0$, we get $t(2-t) \geq 0$, and thus for the sought $t>0$, it is necessary that $t \leq 2$.

For the inequality (1) to hold for all non-negative numbers $x$, the coefficient of $x^{2}$ cannot be negative, otherwise, for sufficiently large non-negative $x$, we would certainly obtain a negative value on the right side of (1). Therefore, it must be that $t-1 \geq 0$ or $t \geq 1$.

On the other hand, for any $t \in [1,2]$, the coefficients of all three terms of the quadratic trinomial (in the variable $x$) on the right side of inequality (1) will be non-negative, so the given inequality will hold for any non-negative real number $x$.

For a complete solution, award 6 points.

For proving that $t \geq 1$, award 2 points (for example, by contradiction: for $t<1$, we rearrange to $x^{2} \geq (t-2) x + t(t-2) / (t-1)$, so the graph of the line represented by the expression on the right side will intersect the $y$-axis at a positive point, which contradicts the position of the vertex of the parabola $y=x^{2}$.

Award 2 points for proving that the inequality is indeed satisfied by the interval $[1,2]$ and for formulating the conclusion.

Deduct 1 point if the mention of equivalent transformations, if necessary, is omitted.

Do not award any points for merely verifying that the statement works for a finite number of $t$ from the interval $[1,2]$.