## Task A-2.1.

If $x, y, z$ and $w$ are real numbers such that

$$
x^{2}+y^{2}+z^{2}+w^{2}+x+3 y+5 z+7 w=4
$$

determine the maximum possible value of the expression $x+y+z+w$.
## Solution.

By completing the square, the given equation can be written in the form

$$
\left(x+\frac{1}{2}\right)^{2}+\left(y+\frac{3}{2}\right)^{2}+\left(z+\frac{5}{2}\right)^{2}+\left(w+\frac{7}{2}\right)^{2}=25
$$

By the inequality between the arithmetic and quadratic means, we have

$$
\frac{\left(x+\frac{1}{2}\right)+\left(y+\frac{3}{2}\right)+\left(z+\frac{5}{2}\right)+\left(w+\frac{7}{2}\right)}{4} \leqslant \sqrt{\frac{\left(x+\frac{1}{2}\right)^{2}+\left(y+\frac{3}{2}\right)^{2}+\left(z+\frac{5}{2}\right)^{2}+\left(w+\frac{7}{2}\right)^{2}}{4}}
$$

We conclude that $\left(x+\frac{1}{2}\right)+\left(y+\frac{3}{2}\right)+\left(z+\frac{5}{2}\right)+\left(w+\frac{7}{2}\right) \leqslant 10$, that is,

$$
x+y+z+w \leqslant 2 \text {. }
$$

Notice that the value 2 is indeed achieved, for example, for $x=2, y=1, z=0, w=-1$.

Therefore, the maximum possible value of the expression $x+y+z+w$ is 2.