26. (ITA) For every integer $n \geq 2$ determine the minimum value that the sum $a_{0}+a_{1}+\cdots+a_{n}$ can take for nonnegative numbers $a_{0}, a_{1}, \ldots, a_{n}$ satisfying the condition
$$
a_{0}=1, \quad a_{i} \leq a_{i+1}+a_{i+2} \quad \text { for } i=0, \ldots, n-2
$$
26. Let us first examine the case that all the inequalities in the problem are actually equalities. Then $a_{n-2}=a_{n-1}+a_{n}, a_{n-3}=2 a_{n-1}+a_{n}, \ldots, a_{0}=$ $F_{n} a_{n-1}+F_{n-1} a_{n}=1$, where $F_{n}$ is the $n$th Fibonacci number. Then it is easy to see (from $F_{1}+F_{2}+\cdots+F_{k}=F_{k+2}$ ) that $a_{0}+\cdots+a_{n}=$ $\left(F_{n+2}-1\right) a_{n-1}+F_{n+1} a_{n}=\frac{F_{n+2}-1}{F_{n}}+\left(F_{n+1}-\frac{F_{n-1}\left(F_{n+2}-1\right)}{F_{n}}\right) a_{n}$. Since $\frac{F_{n-2}\left(F_{n+2}-1\right)}{F_{n}} \leq F_{n+1}$, it follows that $a_{0}+a_{1}+\cdots+a_{n} \geq \frac{F_{n+2}-1}{F_{n}}$, with equality holding if and only if $\alpha_{n}=0$ and $a_{n-1}=\frac{1}{\gamma_{n}}$.
We denote by $M_{n}$ the required minimum in the general case. We shall prove by induction that $M_{n}=\frac{P_{n+2}-1}{F_{n}}$. For $M_{1}=1$ and $M_{2}=2$ it is easy to show that the formula holds; hence the inductive basis is true. Suppose that $n>2$. The sequences $1, \frac{a_{2}}{a_{1}}, \ldots, \frac{a_{0}}{a_{1}}$ and $1, \frac{a}{a_{2}}, \ldots, \frac{a_{n}}{a_{2}}$ also satisfy the conditions of the problem. Hence we have
$$
a_{0}+\cdots+a_{n}=a_{0}+a_{1}\left(1+\frac{a_{2}}{a_{1}}+\cdots+\frac{a_{n}}{a_{1}}\right) \geq 1+a_{1} M_{n-1}
$$
and
$$
a_{0}+\cdots+a_{n}=a_{0}+a_{1}+a_{2}\left(1+\frac{a_{3}}{a_{2}}+\cdots+\frac{a_{n}}{a_{2}}\right) \geq 1+a_{1}+a_{2} M_{n-2} .
$$

Multiplying the first inequality by $M_{n-2}-1$ and the second one by $M_{n-1}$, adding the inequalities and using that $a_{1}+a_{2} \geq 1$, we obtain ( $M_{n-1}+$ $\left.M_{n-2}+1\right)\left(a_{0}+\cdots+a_{n}\right) \geq M_{n-1} M_{n-2}+M_{n-1}+M_{n-2}+1$, so
$$
M_{n} \geq \frac{M_{n-1} M_{n-2}+M_{n-1}+M_{n-2}+1}{M_{n-1}+M_{n-2}+1} .
$$

Since $M_{n-1}=\frac{P_{n+1}-1}{Y_{n-1}}$ and $M_{n-2}=\frac{F_{n}-1}{F_{n-2}}$, the above inequality easily yields $M_{n} \geq \frac{F_{n+2}-1}{Y_{n}}$. However, we have shown above that equality can occur; hence $\frac{F_{v+2}-1}{P_{n}}$ is indeed the required minimum.