2. Determine all pairs $(x, y)$ of real numbers that satisfy the inequality

$$
(x+y)\left(\frac{1}{x}+\frac{1}{y}\right) \geqq\left(\frac{x}{y}+\frac{y}{x}\right)^{2} .
$$
SOLUTION. Neither $x$ nor $y$ can be zero, as they are in the denominators of the fractions in the given inequality. The problem does not specify the sign of the numbers $x$ and $y$, so we must not forget that these numbers can also be negative when performing transformations.

We will first try to simplify the expressions in the problem. By multiplying the inequality by the positive number $x^{2} y^{2}$, we can equivalently eliminate the denominators:

$$
\begin{aligned}
& (x+y) \frac{x+y}{x y} \geqq\left(\frac{x^{2}+y^{2}}{x y}\right)^{2}, \quad \mid \cdot x^{2} y^{2} \\
& x y(x+y)^{2} \geqq\left(x^{2}+y^{2}\right)^{2} .
\end{aligned}
$$

Now it is clear that by expanding the expressions in the obtained inequality, we will get the term $2 x^{2} y^{2}$ on both sides - we can cancel this term in the first step and further transform the inequality into a product form:

$$
\begin{aligned}
x^{3} y+2 x^{2} y^{2}+x y^{3} & \geqq x^{4}+2 x^{2} y^{2}+y^{4}, \\
x^{3} y+x y^{3} & \geqq x^{4}+y^{4}, \\
0 & \geqq x^{4}-x^{3} y+y^{4}-x y^{3}, \\
0 & \geqq x^{3}(x-y)-y^{3}(x-y), \\
0 & \geqq\left(x^{3}-y^{3}\right)(x-y) .
\end{aligned}
$$

On the right side of the last inequality, we have the product of two binomials, and since we are comparing it to zero, it is sufficient to examine the signs of the individual parentheses.

If $x \geqq y$, then $x^{3} \geqq y^{3}$, and similarly, if $x \leqq y$, then $x^{3} \leqq y^{3}$ (this is a consequence of the fact that the function of the third power is increasing over the entire domain of real numbers). Therefore, the expressions in the parentheses have the same sign for any values of $x$ and $y$, so the inequality $\left(x^{3}-y^{3}\right)(x-y) \geqq 0$ holds. Inequality (1) can only be satisfied if one of the parentheses is zero. The equations $x=y$ and $x^{3}=y^{3}$ are equivalent, so inequality (1) holds if and only if $x=y$.

The solution consists of all pairs of real numbers $(x, y)$ where $x=y \neq 0$. In the transformations, we used only equivalent transformations, so a verification of correctness is not necessary.

Note. The product on the right side of (1) can also be transformed using the formula $x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$ into the product of two non-negative polynomials:

$$
\left(x^{3}-y^{3}\right)(x-y)=(x-y)^{2}\left(x^{2}+x y+y^{2}\right) .
$$

The trinomial in the second parenthesis has a non-positive discriminant $D(y)=y^{2}-4 y^{2}=-3 y^{2} \leqq 0$ for any $y$. Therefore, the equality $x^{2}+x y+y^{2}=0$ occurs only in the case $x=y=0$.

## GUIDING AND SUPPLEMENTARY PROBLEMS:

N1. Determine all pairs $(x, y)$ of positive real numbers that satisfy the inequality $x / y + y / x \leqq 2$. [After eliminating the fractions, we transform the inequality to $(x-y)^{2} \leqq 0$, which holds only for $x=y$.]

N2. Determine all pairs $(x, y)$ of real numbers that satisfy the inequality $x / y + y / x \leqq 2$. [The solution consists of all pairs $(x, y)$ such that $x=y \neq 0$ or $x y<0$.]

N3. Determine all pairs $(x, y)$ of real numbers that satisfy the inequality $x^{2}+4 y^{2} \leqq 4 x y$. [The solution consists of all pairs $(x, y)$ such that $x=2 y$.]

N4. Prove that for any real numbers $x, y$, the inequality $x^{4}+y^{4} \geqq x^{3} y+x y^{3}$ holds. [Transformation into a product.]

D1. Prove that for any positive numbers $a, b$, the inequality

$$
\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}} \leqq \sqrt[3]{2(a+b)\left(\frac{1}{a}+\frac{1}{b}\right)}
$$

holds. Determine when equality occurs. [49-A-II-3]