2 Q vs K endgame

Is this statement true?
Claim: There is a #3 (checkmate in 3 moves) in every 2 Q vs K ending if we assume the defending king is between 2 Queens (rank-wise and file-wise) and the attacking King is outside the queens' box( the smallest rectangle containing all the other pieces).
Please write your proof if you can prove it.
note: It's a real question and I started to develop it in my mind after analyzing this game (https://lichess.org/AjlannpgNpKx) 

sample position (black to move)

just google it lol. Find your proof there

You think it's true then?

bro just google it, i won’t do it for you pleaseee

The claim might be true, but honestly this topic is so trivial, what does it matter?

I know it's trivial but yet it's a nice puzzle to think about. It shows how dominant can queens be on open boards. 

I actually proved it true.
To understand the proof first we should understand there are two fundamental checkmates with two queens without the help of any other piece. 
1. 

which is an extension of the swallow's tail mate in a sense. Visual pattern: queens form a 3 x 3 box and the king is in the box. 
2. 

which is an extension of Dovetail's mate in a sense where one Q cuts off two squares of the diagonal and the other one deliver #. Visual pattern: two queens form a finite line with the length of 4 and the king is in touch with the line of queens. The queens dominate a whole 4 x 3 box which is the reason the mirror position is also #.

Since these checkmates are never named anywhere we can call them A and B!
Now the proof: 
if the rectangle containing the two queens is

- 3 × 5 it's #1 in 2 different ways- both of them are of type A. DANGER: With the defensive side to move its stalemate!! 

However, the existence of such a position with the attacking side to move is questionable. 

- 3 × 6 it's #1 in 2 different ways: 1 type A and 1 type B 

- 3 × 7 it's at most #2 there is a possibility of #1 as well.  proof: reduces to 3 × 6 by moving one of the queens one square towards the K while maintaining the width of 3 for the rectangle.

- 3 × 8 it's #2 reduces to 3 × 6 by moving one queen 2 squares towards the K and it's always possible because of the pigeon hole theorem (if the king is distant less than 3 from both queens then the length of the rectangle can't be 8, it can be at most 7)

- 4 × 4 it's #1 in 4 different ways - 2 of type A and 2 of type B. DANGER: With the defensive side to move it's stalemate! Also like the 3 x 5 one, the existence of such a position for the attacking side is impossible! 

- 4 × 5 it's #1 only one checkmate and it's of type B. But it's hypothetical like the previous one and it's not possible to get into such a position. 

- 4 × 6, 4×7, 4 × 8 it's #2 and to understand the technique, we should take a look at this position

I have highlighted the potential squares for the defending K. In all rectangles with width of 4 (4 x 6, 4 x 7 and 4 x 8)  it's the case that there will be two squares on the two corners of the rectangle that there is a #1 of type B.

For all the other squares in the middle, the queens are distinct from each other and one of them has the latitude of 2 from the defending K and the other one has the latitude of 1. The #2 starts with a check by the queen whose  distance from the K is 2 on the intersecting square of the K's vertical line and Q's horizontal line (if the rectangle was vertical, like let's say 8 x 4, vice versa) 

The K will have at most 2 legal moves and it doesn't matter where it goes. The final # will be the same by moving the other Q and delivering a # of type B.

To understand why 4 x 6, 4 x 7 and 4 x 8 are not different and this method works on all of them you can practice more on your own. 
The rest of the proof doesn't have much instructional value but in order to prove the claim I'm going to write it. 

5 x 5, 5 x 6, 5 x 7 and 5 x 8 can be reduced to 4 x 5, 4 x 6, 4 x 7 and 4 x 8 respectively by moving one queen 1 square to the side. 
6 x 6, 6 x 7 and 6 x 8 can be reduced to 4 x 6, 4 x 7 and 4 x 8 respectively. To show why this is always possible we have to do some work
consider the 6 x 6 case, there are 12 potential squares for the K

if the king is too far away from the center one queen can move two squares and reduce the rectangle to width of 4 so we're done there

so it seems the two beset squares for the K might be the central ones

but this one is no different and a too centeralized K can backfire. 

6 x 7, 6 x 8, 7 x 7 and 7 x 8 are similar. The K can't stretch enough on all 4 sides to stop both Qs from cutting the K in a rectangle of width 4. Thus, there is always a #3 by reducing to boxes of width 4. Notice 8 x 8 is not possible since we assumed the attacking K should be outside the Box of other pieces. 

The benefit of knowing this besides the fun of it is in speed chess and by putting the enemy king in a narrow rectangle of width 4 or 3, you can checkmate them in 2 moves. So it's very practical knowledge. But as mentioned above, there are some dangers as well, for example, 3 × 5, 4 × 4, and 4 × 5 boxes are stalemates and you have to be careful. The best way I can think of approaching it is putting the king in narrow and long rectangles since the only stalemates(I have mathematical proof the only ones are the 3 scenarios mentioned above) are occurred in rectangles similar to a square. 

I like it