Prove by induction. If $a_1\geq a_2,a_2\geq a_3,...a_{n-1}\geq a_n$, then $a_1\geq a_n$ with equality iff all a's are equal.
My attempt at a solution:
Suppose n=2, if $a_1\geq a_2$. Clearly then $a_1 \geq a_2$, with equality if $a_1=a_2$ is true.
This is the step I'm not sure about
Then assume $a_1\geq a_{n-1}$, which implies that $a_1-a_{n-1}\geq0$. With equality when $a_1=a_{n-1}$
Also $a_{n-1}\geq a_n$ is given which implies that $a_{n-1}-a_n\geq 0$
We can then add these 2 inequalities $(a_1-a_{n-1})+(a_{n-1}-a_n) \geq 0$
We find that $a_1-a_n\geq 0$
Therefore, by induction
$a_1\geq a_n$, with equality iff all a's are equal.
I'm not sure if the second step works as I believe the property I'm trying to prove isn't really that $a_1 \geq a_n$ but that $a_1\geq a_2$ and $a_2\geq a_3$ implies $a_1\geq a_3$ and I assumed this property in the second step. 
Appreciate any help/critique of my technique here.
 A: I'll use '>' instead of using '≥', and also won't worry about sub-scripting, and use capitalization instead.
An induction step assumes that if a proposition predicated of n holds, then the same proposition predicated of the successor member S(n) holds also.  Or in other words, P(n) implies that P(S(n)), or "if P(n), then P(S(n))".  Here, the successor of n is just (n + 1).  
Now, what is the proposition predicated of n here?
The proposition is the entire if-then statement that you posted.
Thus, changing from using the variable 'n' to 'k' to avoid confusion between the proposition itself and if-then the induction step, the induction step can get written as:
If "If A1 > A2, ... , A(k - 1) > Ak, then A1 > Ak" [this entire if-then part I call the first hypothesis], then "if A1 > A2 ... Ak > A(k + 1) [only the if  part of this I call the second hypothesis], then A1 > A(k + 1)".
Now, by the second hypothesis A1 > A2, ..., A(k - 1) > Ak.  Thus, by detachment and the first hypothesis, A1 > Ak.  Ak > A(k + 1) also holds by the second hypothesis.  Thus, (A1 - Ak) > 0 and Ak > A(k + 1).  
Lastly, adding the inequalities eventually yields A1 > A(k + 1).
So, your technique had potential here.
