Auswahlklausur

Solution:

The function $f:(0, \infty) \rightarrow \mathbb{R}, f(x)=\frac{x+2}{(x+1)(x+5)}$ is strictly monotonically decreasing and convex, since the first derivative
$$f^{\prime}(x)=-\frac{x^{2}+4 x+7}{(x+1)^{2}(x+5)^{2}}$$
is negative, and the second derivative
$$f^{\prime \prime}(x)=\frac{2\left(x^{3}+6 x^{2}+21 x+32\right)}{(x+1)^{3}(x+5)^{3}}$$
is positive. Jensen's inequality gives, for the mean of the arguments $b, c, a$ weighted by the weights $a+1, b+1, c+1$:
$$\frac{(a+1) f(b)+(b+1) f(c)+(c+1) f(a)}{(a+1)+(b+1)+(c+1)} \geq f\left(\frac{(a+1) b+(b+1) c+(c+1) a}{(a+1)+(b+1)+(c+1)}\right)$$
Multiplying by the denominator of the left-hand side, we obtain
$$\frac{(a+1)(b+2)}{(b+1)(b+5)}+\frac{(b+1)(c+2)}{(c+1)(c+5)}+\frac{(c+1)(a+2)}{(a+1)(a+5)} \geq(a+b+c+3) f\left(\frac{(a+1) b+(b+1) c+(c+1) a}{a+b+c+3}\right)$$
Now
$$\begin{aligned} 2((a+1) b+(b+1) c+(c+1) a) & =(a+b+c+3)^{2}-4(a+b+c)-\left(a^{2}+b^{2}+c^{2}\right)-9 \\ & \leq(a+b+c+3)^{2}-4(a+b+c+3) \end{aligned}$$
(in the last step the assumption $a^{2}+b^{2}+c^{2} \geq 3$ was used), hence
$$\frac{(a+1) b+(b+1) c+(c+1) a}{a+b+c+3} \leq \frac{a+b+c-1}{2}$$
Using the monotonicity of $f$, the right-hand side of (*) can thus be further estimated as
$$\geq(a+b+c+3) f\left(\frac{a+b+c-1}{2}\right)$$
This expression now depends only on $s=a+b+c$:
$$=(s+3) \cdot \frac{\frac{s-1}{2}+2}{\left(\frac{s-1}{2}+1\right)\left(\frac{s-1}{2}+5\right)}=\frac{2(s+3)^{2}}{(s+1)(s+9)}=2-\frac{8}{10+s+\frac{9}{s}} \geq 2-\frac{8}{10+6}=\frac{3}{2}$$
(in the second-to-last step the inequality between the arithmetic and geometric mean was applied in the form $s+\frac{9}{s} \geq 2 \sqrt{s \cdot \frac{9}{s}}=6$).