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.

First rewrite the first restriction to $(a-1)^2+(b-1)^2 = 1^2$ and the second restriction to $(c-6)^2+(d-6)^2=6^2$. Hence we may choose $\phi,\theta\in[0,2\pi)$ such that $a = 1+\sin(\phi)$, $b=1+\cos(\phi)$, $c=6+6\sin(\theta)$ and $d = 6+6\cos(\theta)$.
Now $\sin(x)+\cos(x) = \sqrt{2}\sin\left(x+\frac{\pi}{4}\right) \in \left[-\sqrt{2},\sqrt{2}\right]$, hence $a+b = 2+\sin(\phi)+\cos(\phi)\in\left[2-\sqrt{2},2+\sqrt{2}\right]$ and $c+d = 12+6\sin(\theta)+6\cos(\theta)\in\left[12-6\sqrt{2},12+6\sqrt{2}\right]$.
Finally we get $a+b+6(c+d) \in\left[74-37\sqrt{2},74+37\sqrt{2}\right]$.
