Vietnamese Mathematical Competitions

Firstly, we will prove inequality (*) for $n = 4, 5, 6, 7, 8$.

- For $n = 4$, we need to prove $\left(\sum_{i=1}^{4} x_i\right)^2 \ge \frac{4}{3} \sum_{i=1}^{4} x_i (x_{i+1} + x_{i+2} + x_{i+3})$.
We have $\sum_{i=1}^{4} x_i (x_{i+1} + x_{i+2} + x_{i+3}) = \left(\sum_{i=1}^{4} x_i\right)^2 - 4 \sum_{i=1}^{4} x_i^2$ so we need to prove that
$$3\left(\sum_{i=1}^{4} x_i\right)^2 \ge 4\left(\sum_{i=1}^{4} x_i\right)^2 - 4\sum_{i=1}^{4} x_i^2 \Leftrightarrow 4\sum_{i=1}^{4} x_i^2 \ge \left(\sum_{i=1}^{4} x_i\right)^2.$$
The last inequality is true by Cauchy-Schwarz inequality.
So the case $n = 4$ is proved.
The equality occurs if and only if $x_1 = x_2 = x_3 = x_4$.

- For $n = 5$, we should have $\left(\sum_{i=1}^{5} x_i\right)^2 \ge \frac{5}{3} \sum_{i=1}^{5} x_i (x_{i+1} + x_{i+2} + x_{i+3})$.
We have $\sum_{i=1}^{5} x_i (x_{i+1} + x_{i+2} + x_{i+3}) = \left(\sum_{i=1}^{5} x_i\right)^2 - \sum_{i=1}^{5} (x_i^2 + x_i x_{i-1})$ where $x_0 = x_5$ so we need to prove that
$$3\left(\sum_{i=1}^{5} x_i\right)^2 \ge 5\left(\sum_{i=1}^{5} x_i\right)^2 - 5\sum_{i=1}^{5} (x_i^2 + x_i x_{i-1}) \Leftrightarrow 5\sum_{i=1}^{5} (x_i^2 + x_i x_{i-1}) \ge 2\left(\sum_{i=1}^{5} x_i\right)^2 \\ \Leftrightarrow 5\sum_{i=1}^{5} \left(\frac{x_i + x_{i-1}}{2}\right)^2 \ge \left(\sum_{i=1}^{5} x_i\right)^2 \Leftrightarrow 5\sum_{i=1}^{5} \left(\frac{x_i + x_{i-1}}{2}\right)^2 \ge \left(\sum_{i=1}^{5} \frac{x_i + x_{i-1}}{2}\right)^2.$$
The last inequality is true by Cauchy-Schwarz inequality.
The equality occurs if and only if $x_1 = x_2 = x_3 = x_4 = x_5$.

- For $n=6$, we should have $\left(\sum_{i=1}^{6} x_i\right)^2 \ge 2 \sum_{i=1}^{6} x_i (x_{i+1} + x_{i+2} + x_{i+3})$.
We have $\sum_{i=1}^{6} x_i (x_{i+1} + x_{i+2} + x_{i+3}) = 2 \left( \sum_{1 \le i < j \le 6} x_i x_j + x_1 x_4 + x_2 x_5 + x_3 x_6 \right)$ so we need to
prove that
$$\left(\sum_{i=1}^{6} x_i\right)^2 \ge 2 \left( \sum_{1 \le i < j \le 6} x_i x_j + x_1 x_4 + x_2 x_5 + x_3 x_6 \right) \Leftrightarrow (x_1 - x_4)^2 + (x_2 - x_5)^2 + (x_3 - x_6)^2 \ge 0.$$
The last inequality is true. The equality holds if and only if $x_1 = x_4, x_2 = x_5, x_3 = x_6$.

- For $n=7$, we need to prove that $\left(\sum_{i=1}^{7} x_i\right)^2 \ge \frac{7}{3} \sum_{i=1}^{7} x_i (x_{i+1} + x_{i+2} + x_{i+3})$.
We have
$$2 \sum_{i=1}^{7} x_i (x_{i+1} + x_{i+2} + x_{i+3}) = \sum_{i=1}^{7} x_i (x_{i+1} + x_{i+2} + x_{i+3} + x_{i-1} + x_{i-2} + x_{i-3}) = \left(\sum_{i=1}^{7} x_i\right)^2 - \sum_{i=1}^{7} x_i^2$$
So, we need to prove that $\left(\sum_{i=1}^{7} x_i\right)^2 \ge \frac{7}{6} \left[ \left(\sum_{i=1}^{7} x_i\right)^2 - \sum_{i=1}^{7} x_i^2 \right] \Leftrightarrow 7 \sum_{i=1}^{7} x_i^2 \ge \left(\sum_{i=1}^{7} x_i\right)^2$.
The last inequality is true by Cauchy-Schwarz inequality.
The equality occurs if and only if $x_1 = x_2 = x_3 = x_4 = x_5 = x_6 = x_7$.

- For $n=8$, we should have $\left(\sum_{i=1}^{8} x_i\right)^2 \ge \frac{8}{3} \sum_{i=1}^{8} x_i (x_{i+1} + x_{i+2} + x_{i+3})$. We have
$$2 \sum_{i=1}^{8} x_i (x_{i+1} + x_{i+2} + x_{i+3}) = \sum_{i=1}^{8} x_i (x_{i+1} + x_{i+2} + x_{i+3} + x_{i-1} + x_{i-2} + x_{i-3}) = \left(\sum_{i=1}^{8} x_i\right)^2 - \sum_{i=1}^{8} (x_i^2 + x_i x_{i+4})$$
We need to prove that
$$\begin{aligned} 3 \left(\sum_{i=1}^{8} x_i\right)^2 &\ge 4 \left(\sum_{i=1}^{8} x_i\right)^2 - 4 \sum_{i=1}^{8} (x_i^2 + x_i x_{i+4}) \\ &\Leftrightarrow 4 \sum_{i=1}^{8} (x_i^2 + x_i x_{i+4}) \ge \left(\sum_{i=1}^{8} x_i\right)^2 \\ &\Leftrightarrow 8 \sum_{i=1}^{8} \left(\frac{x_i + x_{i+4}}{2}\right)^2 \ge \left(\sum_{i=1}^{8} \frac{x_i + x_{i+4}}{2}\right)^2 \end{aligned}$$
The last inequality is true by Cauchy-Schwarz inequality. The equality occurs iff
$x_1 + x_5 = x_2 + x_6 = x_3 + x_7 = x_4 + x_8$.

Next, for $n \ge 8$, we will prove that $3\left(\sum_{i=1}^{n} x_i\right)^2 \ge 8\sum_{i=1}^{n} x_i (x_{i+1} + x_{i+2} + x_{i+3})$ by induction.
- For $n=8$ the statement is true.
- Assume that the statement is true for all $n$-tuple $(x_1, x_2, x_3, \dots, x_n)$. Consider $n+1$ nonnegative numbers $(x_1, x_2, x_3, \dots, x_n, x_{n+1})$ with $x_{n+1} = \min\{x_1, x_2, x_3, \dots, x_n, x_{n+1}\}$.
Put $S_n = \sum_{i=1}^{n} x_i$, $P_n = \sum_{i=1}^{n} x_i (x_{i+1} + x_{i+2} + x_{i+3})$, we have to prove that $S_n^2 \ge \frac{8}{3}P_n$ for $n \ge 8$.
We have
$$3S_{n+1}^2 = 3(S_n + x_{n+1})^2 = 3(S_n^2 + x_{n+1}^2 + 2x_{n+1}S_n) \text{ and}$$
$$8P_{n+1} = 8P_n + 8[x_{n-2}(x_{n+1}-x_1) + x_{n-1}(x_{n+1}-x_2) + x_n(x_{n+1}-x_3) + x_{n+1}(x_1+x_2+x_3)].$$
By the induction hypothesis then $S_n^2 \ge \frac{8}{3}P_n$ so, it is sufficient to prove that
$$3x_{n+1}^2 + 6x_{n+1}S_n \ge 8[x_{n-2}(x_{n+1}-x_1) + x_{n-1}(x_{n+1}-x_2) + x_n(x_{n+1}-x_3) + x_{n+1}(x_1+x_2+x_3)] \text{ (**)}$$
We have $(x_1 - x_{n+1})(x_{n-2} - x_{n+1}) \ge 0 \Rightarrow x_{n+1}^2 - x_1 x_{n+1} \ge x_{n-2}(x_{n+1} - x_1)$. Similarly, we construct the other inequalities and add them up, we have
$$3x_{n+1}^2 \ge x_{n-2}(x_{n+1}-x_1) + x_{n-1}(x_{n+1}-x_2) + x_n(x_{n+1}-x_3) + x_{n+1}(x_1+x_2+x_3)$$
So, it suffices to prove that
$$3x_{n+1}^2 + 6x_{n+1}S_n \ge 24x_{n+1}^2 \Leftrightarrow x_{n+1}(2S_n - 7x_{n+1}) \ge 0.$$
Since $n \ge 4$ we have $2S_n \ge 8x_{n+1} \ge 7x_{n+1}$ and the last inequality is true.
Thus, the statement is true for $n+1$. By induction principle, the inequality is true for all $n \ge 8$. So (*) holds for all $n$.

b) We find the condition for equality in the inequality.
For $n=4$, equality holds if and only if $x_1 = x_2 = x_3 = x_4$.
For $n=5$, equality holds if and only if $x_1 = x_2 = x_3 = x_4 = x_5$.
For $n=6$, equality holds if and only if $x_1 = x_4, x_2 = x_5, x_3 = x_6$.
For $n=7$, equality holds if and only if $x_1 = x_2 = x_3 = x_4 = x_5 = x_6 = x_7$.
For $n=8$, equality holds if and only if $x_1 + x_5 = x_2 + x_6 = x_3 + x_7 = x_4 + x_8$.
Consider $n \ge 9$, by the proof by induction above, the equality holds if and only if
$$x_{n+1} = x_{n-2}x_1 = x_{n-1}x_2 = x_n x_3 = 0 \quad \text{and} \quad S_n^2 = \frac{8}{3}P_n \quad \text{with } (***)$$
If $n=9$ we should have the equality for $n$ first numbers, as well as we should have $x_9 = x_6x_1 = x_7x_2 = x_8x_3 = 0$, that is
$$\begin{cases} x_9 = x_6x_1 = x_7x_2 = x_8x_3 = 0 \\ x_1 + x_5 = x_2 + x_6 = x_3 + x_7 = x_4 + x_8 \end{cases}$$
If $x_2 = x_6 = 0$ or $x_3 = x_7 = 0$ then it is easy to see that all other numbers equal to 0; if not all $x_2, x_6$ and $x_3, x_7$ are 0, we can assume that $x_1 = x_2 = x_3 = 0$, which leads to the conclusion that the tuple that satisfies the system of conditions is $(0,0,0,0,a,a+b,a+b,a+b,b)$ and permutations. It is easy to see that the case when all numbers are zero can be included in this case.
If $n \ge 9$ then the equality for $S_{n+1}^2 \ge \frac{8}{3}P_{n+1}$ requires conditions (****). By induction, we can prove that the equality holds only for the tuple
$$(\underbrace{0,0,0,\dots,0}_{n-5}, a, a+b, a+b, a+b, b) \text{ and its permutations.}$$