## 3. Consider the expression

$$
2 x^{2}+y^{2}-2 x y+2 x+4 \text {. }
$$

a) Find all real numbers $x$ and $y$ for which the given expression attains its minimum value.

b) Determine all pairs of non-negative integers $x$ and $y$ for which the value of the given expression is equal to 16.
SOLUTION. Let's transform the given expression $V(x, y)$ using the formulas for $(A \pm B)^{2}$:

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

a) The first two addends in the last sum are squares, so they have non-negative values. The minimum certainly occurs when for some $x$ and $y$ both bases are zero (in that case, for another pair of bases, the value of the expression $V(x, y)$ will be greater). Both equalities $x-y=0, x+1=0$ indeed occur simultaneously, and clearly only for the values $x=y=-1$. Let's add (though the problem does not ask for it) that $V_{\min }=V(-1,-1)=3$.

Answer. The given expression attains its minimum value only for $x=y=-1$.

b) According to the transformation at the beginning of the solution, we have

$$
V(x, y)=16 \Leftrightarrow(x-y)^{2}+(x+1)^{2}+3=16 \Leftrightarrow(x-y)^{2}+(x+1)^{2}=13 .
$$

Both addends $(x-y)^{2}$ and $(x+1)^{2}$ (for non-negative integers $x$ and $y$) are from the set $\{0,1,4,9,16, \ldots\}$. Therefore, one of them must be 4 and the other 9. Given the assumption $x \geqq 0$, the base $x+1$ of the power $(x+1)^{2}$ is positive, so it must be 2 or 3 (and not -2 or -3). In the first case, i.e., for $x=1$, we get the condition $1-y= \pm 3$, i.e., $y=1 \mp 3$ or $y=4$ (the value $y=-2$ is excluded by the condition in part b). In the second case, when $x=2$, we similarly get from the equality $x-y=2-y= \pm 2$ two suitable values $y=0$ and $y=4$.

Answer. All the sought pairs $(x, y)$ are $(1,4),(2,0)$ and $(2,4)$.

GUIDING AND SUPPLEMENTARY PROBLEMS:

N1. For any real numbers $x, y$ and $z$, prove the non-negativity of the values of each of the expressions $x^{2} z^{2}+y^{2}-2 x y z, \quad x^{2}+4 y^{2}+3 z^{2}-2 x-12 y-6 z+13, \quad 2 x^{2}+4 y^{2}+z^{2}-4 x y-2 x z$ and also determine when the corresponding value is zero.

D1. Solve the equation $x^{2}+y^{2}+x+y=4$ in the set of integers. [B61-S-1]

D2. For positive real numbers $a, b, c$ it holds that $c^{2}+a b=a^{2}+b^{2}$. Show that then it also holds that $c^{2}+a b \leqq a c+b c$. [C63-II-3]

D3. Prove that for any non-negative numbers $a, b, c$ it holds that $(a+b c)(b+a c) \geqq a b(c+1)^{2}$. Determine when equality occurs. [C58-S-1]

D4. Consider the expression $V(x)=\left(5 x^{4}-4 x^{2}+5\right) /\left(x^{4}+1\right)$.

a) Prove that for every real number $x$ it holds that $V(x) \geqq 3$.

b) Find the maximum value of $V(x)$. [C58-II-1]

D5. Prove that for any distinct positive numbers $a, b$ it holds that

$$
\frac{a+b}{2}<\frac{2\left(a^{2}+a b+b^{2}\right)}{3(a+b)}<\sqrt{\frac{a^{2}+b^{2}}{2}} .
$$

[C58-I-6]