1. a) (3p) Determine the minimum of the expression $E(a)=(a+2)\left(a^{2}-6 a+16\right)$, where $a \in[0,+\infty)$.

b) (4p) Prove that for any positive real numbers $a, b, c \in[0,+\infty)$, with $a+b+c=6$, the inequality holds: $\frac{1}{a^{2}-6 a+16}+\frac{1}{b^{2}-6 b+16}+\frac{1}{c^{2}-6 c+16} \leq \frac{3}{8}$. Specify the case of equality.

Dan Popescu, Suceava
Solution: a) $E(a)=(a+2)\left(a^{2}-6 a+16\right)=a^{3}-6 a^{2}+16 a+2 a^{2}-12 a+32=a^{3}-4 a^{2}+4 a+32=$ $a(a-2)^{2}+32 \geq 32 \Rightarrow \min \{E(a), a \geq 0\}=32$, the minimum which is achieved for $a \in\{0,2\}$.

b) From the previous subpoint we have:

$$
\frac{1}{a^{2}-6 a+16} \leq \frac{a+2}{32}, \frac{1}{b^{2}-6 b+16} \leq \frac{b+2}{32}, \frac{1}{c^{2}-6 c+16} \leq \frac{c+2}{32}, \forall a, b, c \in[0,+\infty)
$$

Adding the three inequalities term by term, we get

$\frac{1}{a^{2}-6 a+16}+\frac{1}{b^{2}-6 b+16}+\frac{1}{c^{2}-6 c+16} \leq \frac{a+2+b+2+c+2}{32}=\frac{12}{32}=\frac{3}{8} . \quad$ Equality holds if $a, b, c \in\{0,2\}$ and from the condition $a+b+c=6$ we get $a=b=c=2$.

## Scoring:

| a) Determine $E(a)=a^{3}-4 a^{2}+4 a+32$ | $1 \text{p}$ |
| :--- | :---: |
| Find the minimum | $2 \text{p}$ |
| b) Write $\frac{1}{a^{2}-6 a+16} \leq \frac{a+2}{32}, \forall a \in[0,+\infty)$ and analogues | $2 \text{p}$ |
| Add inequalities, finalize | $1 \text{p}$ |
| Determine when equality holds | $1 \text{p}$ |