## Task B-2.4.

The sum of four positive real numbers is 8, and the sum of their squares is 40. Determine the maximum possible value that one of these numbers can take.
## Solution.

Let $a, b, c, d$ be the observed numbers. Without loss of generality, we can assume that $d$ is the largest of them. From $a+b+c+d=8$ and $a^{2}+b^{2}+c^{2}+d^{2}=40$ it follows that

$$
a+b+c=8-d, \quad a^{2}+b^{2}+c^{2}=40-d^{2}
$$

If we square the first equation and apply the second, we get

$$
\begin{gathered}
(a+b+c)^{2}=(8-d)^{2} \\
a^{2}+b^{2}+c^{2}+2(a b+b c+c a)=(8-d)^{2} \\
40-d^{2}+2(a b+b c+c a)=64-16 d+d^{2} \\
a b+b c+c a=12-8 d+d^{2}
\end{gathered}
$$

It is known that

$$
a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a
$$

(easily derived from the inequality $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geqslant 0$ ).

If we apply this inequality $(**)$ and $(***)$, we get:

$$
\begin{gathered}
a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a \\
40-d^{2} \geqslant 12-8 d+d^{2} \\
d^{2}-4 d-14 \leqslant 0
\end{gathered}
$$

The solutions to the equation $d^{2}-4 d-14=0$ are $d=2 \pm 3 \sqrt{2}$, and the solution to the above inequality is $d \in[2-3 \sqrt{2}, 2+3 \sqrt{2}]$. The largest real number $d$ in this interval is $2+3 \sqrt{2}$.

We need to check if there exist positive real numbers $a, b$, and $c$ that, along with $d=2+3 \sqrt{2}$, satisfy both given equations.

If we set $a=b=c=\frac{8-d}{3}=2-\sqrt{2}$, it is easy to verify that indeed $a^{2}+b^{2}+c^{2}+d^{2}=$ $3(2-\sqrt{2})^{2}+(2+3 \sqrt{2})^{2}=40$.

The maximum possible value of one of the observed numbers is $2+3 \sqrt{2}$.

Note: Instead of inequality $(***)$, we can apply the inequality between the arithmetic and quadratic means:

$$
\begin{aligned}
& \frac{a+b+c}{3} \leqslant \sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}} \\
& (a+b+c)^{2} \leqslant 3\left(a^{2}+b^{2}+c^{2}\right) \\
& (8-d)^{2} \leqslant 3\left(40-d^{2}\right) \\
& d^{2}-4 d-14 \leqslant 0
\end{aligned}
$$