For what minimum value of $t$ does the inequality $\sqrt{x y} \leq t(2 x+3 y)$ hold for all non-negative real numbers $x, y$?
Solution. In the inequality of the problem, substituting 3 for $x$ and 2 for $y$, it must hold that $t \cdot 12 \geq \sqrt{6}$, and thus the value of $t$ must be at least $\frac{\sqrt{6}}{12}=\frac{1}{2 \sqrt{6}}$. We will show that the inequality holds for all such values of $t$ if $x$ and $y$ are non-negative numbers.

By the inequality between the arithmetic and geometric means, for non-negative $2x$ and $3y$,

$$
2 x+3 y \geq 2 \sqrt{2 x \cdot 3 y}=2 \sqrt{6} \sqrt{x y}
$$

If $t \geq 0$, then multiplying by $t$,

$$
t(2 x+3 y) \geq t \cdot 2 \sqrt{6} \sqrt{x y}
$$

If $t \geq \frac{1}{2 \sqrt{6}}$, that is, $t \cdot 2 \sqrt{6} \geq 1$, then the right-hand side of (1) is greater than or equal to $\sqrt{x y}$, so the inequality in question is true.

The inequality in question, therefore, holds for all non-negative $x, y$ real numbers if and only if $t$ is at least $\frac{1}{2 \sqrt{6}}$. (Aron Kalcsu, Zalaegerszeg, Zrínyi Miklós Gymnasium, 12th grade)

Note. The solutions essentially followed the above path. Of course, any substitution that results in equality in the inequality between the arithmetic and geometric means is suitable in the first step, that is, $2 x=3 y$. Something is simplified by the structure of the problem if we rearrange it so that neither side depends on the variables, for example,

$$
A(x, y)=\frac{\sqrt{x y}}{2 x+3 y} \leq t
$$

In this way, it becomes clearer that we are actually looking for the smallest upper bound of the function $A(x, y)$. The inequality between the arithmetic and geometric means implies that the function $A(x, y)$ has a maximum value, and this maximum value is precisely $\frac{1}{2 \sqrt{6}}$, which is the solution to the problem.