Prove inequality: $74 - 37\sqrt 2 \le a+b+6(c+d) \le 74 +37\sqrt 2$ without calculus 
Let $a,b,c,d \in \mathbb R$ such that $a^2 + b^2 + 1 = 2(a+b),
c^2 + d^2 + 6^2 = 12(c+d)$, prove inequality without calculus (or
  langrange multiplier):  $$74 - 37\sqrt 2  \le a+b+6(c+d) \le 74
 +37\sqrt 2$$

The original problem is find max and min of  $a+b+6(c+d)$ where ...
Using some calculus, I found it, but could you solve it without calculus.

HINT:
So, $(a-1)^2+(b-1)^2=1^2$  and we can set $a=1+\cos A,b=1+\sin A$
So, $a+b=2+(\cos A+\sin A)=2+\sqrt2\cos\left(A-\frac\pi4\right)$
As $-1\le \cos\left(A-\frac\pi4\right)\le 1, 2-\sqrt2\le a+b\le 2+\sqrt2 $
Similarly,  $(c-6)^2+(d-6)^2=6^2$ and we can set $c=6+6\cos B,d=6+6\sin B$
