Maximum number of guaranteed coins to get in a "30 coins in 3 boxes" puzzle 
There are a total of 30 gold coins in three wooden boxes (but you do not know
  how many in each individual box). However, you know that one box has exactly 4 coins more than another box.
For each box, you can ask for a number of coins from that box, of your choice. If there are at least that many coins in that box, then you get as many coins as you asked for. Otherwise, you get nothing from that box. You must place all your demands simultaneously in the beginning.
What is the maximum number of coins that you can guarantee yourself to get?

Added for clarification: you don't know which box has 4 coins more.
Let $x, y, y+4$ be the coin contents $=> x+2y=26$ $=>$ solutions $(0,13,17),...,(24,1,5)$.
What next?
 A: If you ask for $13,13,13$, then you get nothing if the boxes are $10,8,12$.

If you ask for $12,12,12$, you will always get exactly $12$ coins, since in all cases, exactly one of the boxes has at least $12$ coins.

Claim $12$ is best possible.

Certainly it's best possible with all requests equal.

Suppose a better request triple is $a,b,c$, with $a \le b \le $c, and $a < c$.

But any of the $3$ requests might fail if that box contains $0$ coins, hence to be sure to get more than $12$ coins, every pair must sum to more than $12$. 

From $a \le b \le c $, and $a + b > 12$, we get $6 < b \le c$.

But if the boxes are some permutation of $26,0,4$, the $b,c$ requests might both fail, so to be sure to get more than $12$, we must have $a > 12$.

But then $a,b,c$ all exceed $12$, so all requests would fail if the boxes are $10,8,12$.

Thus, unequal requests can't beat the uniform $12,12,12$ request strategy.

It follows that $12$ is the maximum number of coins which can be guaranteed.
A: There is always one box with at least 12 coins. If you name (12, 12, 12), you will win at least 12 coins. 
If you name some triple where one of the numbers is more than 12, then it's possible that box only has 12, and you win nothing from it. If you are to win from both the other boxes, your other two named numbers must be at most 2, since the other boxes could have 2 and 16. So then you'd win at most 4 total. If you are to win from exactly one of the other two boxes, they might have 8 and 10 and you'd win at most 10 total.
If you name some triple where your lowest number is less than 12 but more than 4, then it's possible that box has 26, and you win nothing from the other two boxes which have 0 and 4 coins. And you only win your lowest named number which we assumed is less than 12.
If your lowest called number is 4 or lower, that box may again have 26, and you can't do better than winning that number plus the 4 from the box with 4. So no better than 8 total.
So your lowest named number is at least 12, and none of your named numbers are more than 12. So the best you can do is guarantee 12 coins with (12,12,12). Note we've shown this is the only triple to guarantee 12 won coins.
A: After much thought, I realised that (10, 8, 12) can only be the optimal solution.  Firstly, we have (10, 10, 10) to ensure we get the most coins in a situation with no constraints.  However, here, we have to take note of the 4 coins, so I would say (10, 8, 12).
However, should the boxes with more coins be unknown to us, then alex's answer of (12, 12, 12) would be more suitable.
A: This is not an answer, just in case we know which box has 4 coin more, we can have a better strategy.
Say box 3 has 4 more coin than box 2, label them as you did $(x,\ y,\ y+4)$.
We can demand for $(16, 6, 10)$, so that in any case we get $16$ coins.
