logarithmical inequality with equality case proof If 10$\leq$a$_1$$\leq$a$_2$$\leq$...$\leq$a$_{2020}$$\leq$100 with a$_i$ are real numbers, i=$\overline{1,2020}$ then prove that $$\sum_{i=1}^{2019} log_{a_i}a_{i+1}\leq2020 $$ and the equality case it happens if and only if a$_n$=10 for n$\leq$2000 and $a_n$=100 for n$\geq$2001
my solution:  a$_i$$\geq$10 so lg a$_i$$\geq$1 for i=$\overline{1,2020}$ and the left member becomes $$\sum_{i=1}^{673} \frac{lga_{i+1}}{lga_i}$$ +$$\sum_{i=674}^{2019} log_{a_i}a_{i+1} $$ $\leq$ $$\sum_{i=2}^{674} lga_i$$ + $$\sum_{i=674}^{2019} log_{a_i}a_{i+1} $$ and because log$_{a_i}a_{i+1}$$\leq$log$_{a_i}$100=$\frac{1}{2lga_{2000}}$$\leq$$\frac{1}{2}$ then the inequality becomes lg100$\cdot$673 + $\frac{1}{2}$$\cdot$1346=2019<2020
and i have proved that inequality its strict <2020 but i didn't know how to proof the part with equality case.
 A: It is enough to show that for $1<a\leqslant b\leqslant c$
$$\log_a b + \log_b c \leqslant 1+\log_ac \tag{$\star$}$$
as this will “telescopically bound” the sum to $2018+\log_{a_1}a_{2020} \leqslant 2018+\log_{10}100$.
For $\star$, we have
$$\iff \log b \cdot \log b + \log a \cdot \log c \leqslant \log a \log b + \log b \log c$$
which follows from rearrangement inequality, as $\log$ is increasing, and $(a,b)$ and $(b,c)$ are similarly ordered.
The equality case is iff $\{a_k: k\in [2020]\}=\{10, 100\}$.
A: You have stated the problem incorrectly.  As stated, the minumum is 2019, and this occurs at 
$$
a_1 = a_2 = a_3 = \cdots = a_{2020} = a 
$$ 
for any value $10 \leq a \leq 100$.
I suspect the problem originally stated that 
$$
10 \color{red}= a_1 \leq a_2 \leq a_3 \leq \cdots \leq a_{2020} \color{red}= 100
$$
Then the minimum value is indeed $2020$ but that can be achieved for any set of $a_i$ such that they are in non-decreasing order, with $a_1 = 10, a_{2020}=100, \forall i : a_i \in \{ 10, 100 \}$.
For example, choose $a_1 = 10$ and all other $a_i = 100$ and the sum still works out to $2020$.  
A: I dont see how your proof is correct, but if it is equality you are looking for, take $a_1=10, a_2=...=a_{2020}=100$, and equality follows easily. For the main proof, i would suggest something of the sort:
lower bounding logarithm of sums
