2. What is the greatest possible value of the expression $x y - x^{3} y - x y^{3}$, where $x$, $y$ are positive real numbers? For which $x$, $y$ is this value achieved?

(Mária Dományová, Patrik Bak)
SOLUTION. From the transformation of the given expression $V$ into a product

$$
V=x y-x^{3} y-x y^{3}=x y \cdot\left(1-x^{2}-y^{2}\right)
$$

we can see that if the factor $1-x^{2}-y^{2}$ is negative, the expression $V$ is also negative. Therefore, it suffices to consider only positive $x$ and $y$ for which $1-x^{2}-y^{2}>0$, as is the case for $x=y=\frac{1}{2}$, or more generally for any $x, y$ close to zero.

Thanks to this restriction, we can estimate the value of $V$ from above by first estimating the product $x y$ using the well-known AG-inequality, which is proven in the introductory problem 1. In it, we set $u=x^{2}$ and $v=y^{2}$. We then get the inequality $x y \leqq \frac{1}{2}\left(x^{2}+y^{2}\right)$. Multiplying both sides by the positive number $1-x^{2}-y^{2}$, we obtain the upper bound

$$
V=x y \cdot\left(1-x^{2}-y^{2}\right) \leqq \frac{1}{2}\left(x^{2}+y^{2}\right) \cdot\left(1-\left(x^{2}+y^{2}\right)\right)
$$

If we now set $t=x^{2}+y^{2}$, it remains to find the maximum of the quadratic function $f(t)=t(1-t)$ for $t \in(0,1)$. This maximum can be found by again using the AG-inequality. This time, we choose (positive) values $u=t$ and $v=1-t$, where $t=x^{2}+y^{2}$. Since $u+v=1$, the AG-inequality has the form $\sqrt{t(1-t)} \leqq \frac{1}{2}$, which, after squaring both sides, gives us $t(1-t) \leqq \frac{1}{4}$. Substituting $t=x^{2}+y^{2}$, we get

$$
\left(x^{2}+y^{2}\right) \cdot\left(1-\left(x^{2}+y^{2}\right)\right) \leqq \frac{1}{4}
$$

From the estimate (1), it follows that $V \leqq \frac{1}{2} \cdot \frac{1}{4}=\frac{1}{8}$.

Is the number $\frac{1}{8}$ a possible value of the expression $V$? And if so, for which admissible pairs $(x, y)$ is it achieved? The answer is provided by the previous process. According to it, the equality $V=\frac{1}{8}$ is achieved precisely when equality holds in both applied AG-inequalities. As is known, this happens precisely when $u=v$ for both pairs of numbers $u, v$. In our situation, this means the equalities $x^{2}=y^{2}$ and $t=1-t$ for $t=x^{2}+y^{2}$. The equivalent condition $x^{2}=y^{2}=\frac{1}{4}$ corresponds to the only admissible pair $(x, y)=\left(\frac{1}{2}, \frac{1}{2}\right)$.

Conclusion. The greatest possible value of the given expression is $\frac{1}{8}$. The expression achieves this value for the unique pair $(x, y)=\left(\frac{1}{2}, \frac{1}{2}\right)$.

NOTE. Both inequalities used in the solution

$$
x y \leqq \frac{1}{2}\left(x^{2}+y^{2}\right) \quad \text { and } \quad t(1-t) \leqq \frac{1}{4}
$$

can be easily proven without referring to the AG-inequality by "squaring" transformations, i.e., rewriting them into equivalent forms

$$
\frac{1}{2}(x-y)^{2} \geqq 0, \quad \text { resp. } \quad\left(t-\frac{1}{2}\right)^{2} \geqq 0
$$

From these, the necessary and sufficient conditions for the cases of equality also follow, i.e., $x=y$, resp. $t=\frac{1}{2}$ (as well as the validity of both inequalities for any real numbers $x, y, t$ regardless of their signs).

GUIDING AND ADDITIONAL PROBLEMS:

N1. Prove that for any non-negative real numbers $u$, $v$ the inequality $\sqrt{u v} \leqq \frac{1}{2}(u+v)$ holds, and equality in it occurs precisely when $u=v$. [Indeed, it holds that $(\sqrt{u}-\sqrt{v})^{2} \geqq 0$. Expanding the left side of this inequality, we rewrite it as $u-2 \sqrt{u v}+v \geqq 0$, which we finally rearrange to the required $\sqrt{u v} \leqq \frac{1}{2}(u+v)$. Equality in this inequality occurs precisely when $\sqrt{u}-\sqrt{v}=0$, which is equivalent to $\sqrt{u}=\sqrt{v}$, and thus also $u=v$. NOTE. Since the expression on the left side of the inequality is the geometric mean of two non-negative real numbers $u, v$ and the expression on the right side is their arithmetic mean, the inequality mentioned is called the inequality between the arithmetic and geometric means of two non-negative real numbers, or briefly the AG-inequality. A similarly named inequality holds for $n$-tuples of non-negative numbers.]

N2. Find the maximum value of the expressions a) $t(1-t)$, b) $u v(1-u v)$, c) $\left(u^{2}+v^{2}\right)\left(1-u^{2}-v^{2}\right)$. In all expressions, the letters represent arbitrary real numbers. [a) $\frac{1}{4}$. If both numbers $t$ and 1 - $t$ are positive, write the AG-inequality for them. Consider the cases where both numbers are not positive. An alternative option is to rewrite it as a square: $t(1-t)=\frac{1}{4}-\left(\frac{1}{2}-t\right)^{2}$. b) $\frac{1}{4}$. Substitution $t=u v$ leads to case a). c) $\frac{1}{4}$. Substitution $t=u^{2}+v^{2}$ leads to case a).]

D1. For real numbers $a, b$, find the maximum possible value of the expression

$$
\frac{a b}{a^{2}+b^{2}} \text {. }
$$

[The maximum is $\frac{1}{2}$ (for $a=b$). It is sufficient to consider the case where $a>0$ and $b>0$. Use the AG-inequality for the pair of numbers $a^{2}$ and $b^{2}$. Alternatively, it is possible to start from the inequality $(a-b)^{2} \geqq 0$, rearranged to the form $2 a b \leqq a^{2}+b^{2}$.]

D2. For non-negative real numbers $a, b$ it holds that $a+b=2$. Determine the minimum and maximum possible value of the expression

$$
V=\frac{a^{2}+b^{2}}{a b+1}
$$

$[68-\mathrm{B}-\mathrm{II}-1]$

D3. Prove that for any real numbers $a, b, c$ it holds that $a^{2}+b^{2}+c^{2} \geqq a b+b c+c a$. [By adding three inequalities $a^{2}+b^{2} \geqq 2 a b, b^{2}+c^{2} \geqq 2 b c, c^{2}+a^{2} \geqq 2 c a$, we obtain an inequality which it then suffices to divide by two.]

D4. For non-negative real numbers $a, b$ it holds that $a^{2}+b^{2}=1$. Determine the minimum and maximum possible value of the expression

$$
V=\frac{a^{4}+b^{4}+a b+1}{a+b} .
$$

$[68-\mathrm{B}-\mathrm{I}-4]$