10. Finde die grösste natürliche Zahl $n$, sodass für alle reellen Zahlen $a, b, c, d$ folgendes gilt:

$$
(n+2) \sqrt{a^{2}+b^{2}}+(n+1) \sqrt{a^{2}+c^{2}}+(n+1) \sqrt{a^{2}+d^{2}} \geq n(a+b+c+d)
$$
Solution: The idea is to apply Cauchy-Schwarz to the three roots on the left:

$$
\begin{aligned}
& (n+2) \sqrt{a^{2}+b^{2}}+(n+1) \sqrt{a^{2}+c^{2}}+(n+1) \sqrt{a^{2}+d^{2}} \\
= & \sqrt{(4 n+4)+n^{2}} \sqrt{a^{2}+b^{2}}+\sqrt{(2 n+1)+n^{2}} \sqrt{a^{2}+c^{2}}+\sqrt{(2 n+1)+n^{2}} \sqrt{a^{2}+d^{2}} \\
\geq & (2 a \sqrt{n+1}+n b)+(a \sqrt{2 n+1}+n c)+(a \sqrt{2 n+1}+n d) \\
= & a(2 \sqrt{n+1}+2 \sqrt{2 n+1})+n(b+c+d)
\end{aligned}
$$

For the three inequalities we have applied, there is, for example, the following case of equality

$$
(a, b, c, d)=\left(1, \frac{n}{2 \sqrt{n+1}}, \frac{n}{\sqrt{2 n+1}}, \frac{n}{\sqrt{2 n+1}}\right)
$$

which, by replacing in the starting inequality, gives us the necessary condition :

$$
2(\sqrt{n+1}+\sqrt{2 n+1}) \geq n
$$

Squaring twice and simplifying gives $n \leq 24$ with equality if $n=24$. So $n=24$ is the optimum value.