Murad.Aghazade wrote: $\boxed{\text{A2}}$ Find the maximum value of $|\sqrt{x^2+4x+8}-\sqrt{x^2+8x+17}|$ where $x$ is a real number. Consider in the plane the points $A(-2,2)$, $B(-4,1)$ and $M(0,x)$. Required quantity is $|AM-BM|$ And since $|AM-BM|\le AB$ with equality when $A,B,M$ are on the same line, we get the answer : $\boxed{\text{maximum is }AB=\sqrt 5}$

We can take the derivative of $\sqrt{x^2+4x+8}-\sqrt{x^2+8x+17}$: $$\frac{(x^2+4x+8)’}{2\sqrt{x^2+4x+8}} - \frac{(x^2+8x+17)’}{2\sqrt{x^2+4x+8}} = \frac{2x+4}{2\sqrt{x^2+4x+8}} - \frac{2x+8}{2\sqrt{x^2+4x+8}}$$and since we know optimality will occur at $0$: $$\frac{2x+4}{2\sqrt{x^2+4x+8}} = \frac{2x+8}{2\sqrt{x^2+4x+8}} \Longleftrightarrow \frac{(x+2)^2}{x^2+4x+8} = \frac{(x+4)^2}{x^2+8x+17}$$which can be written as $$x^4+12x^3+53x^2+100x+68 = x^4+12x^3 + 56x^2 + 128x + 128 \Longleftrightarrow 3x^2 + 28x+60 = 0.$$Solving, we see that $x = \frac{-10}{3}, -6$, and it’s clear that $-6$ brings the maximum value of $\boxed{\sqrt{5}}.$ However, we still need to be careful with the edge cases--it’s possible that the maximum value occurs when $x$ approaches $\infty$ or $-\infty$ $$\lim_{ x \to \infty} |\sqrt{x^2+4x+8}-\sqrt{x^2+8x+17}| = \lim_ {x \to \infty} \sqrt{x^2+8x+17}-\sqrt{x^2+4x+8} = \lim_ {x \to \infty} \sqrt{x^2+8x+16}-\sqrt{x^2+4x+4} = (x+4)-(x+2) = 2$$which is clearly less than $\sqrt{5}$. On the other hand, $$\lim_{ x \to -\infty} |\sqrt{x^2+4x+8}-\sqrt{x^2+8x+17}| = \lim_{ x \to -\infty} \sqrt{x^2+4x+8}-\sqrt{x^2+8x+17} = \lim_{ x \to -\infty} \sqrt{x^2+4x+4}-\sqrt{x^2+8x+16} = (-x-2) - (-x-4) = 2$$which is clearly less than $\sqrt{5}$. Hence, $\sqrt{5}$ is the maximum.

For convenience, set $u = x+3$, then we are interested in $\left|\sqrt{(u-1)^2 + 4} - \sqrt{(u+1)^2+1} \right|$. Trial and error leads to the guess that $\left|\sqrt{(u-1)^2 + 4} - \sqrt{(u+1)^2+1} \right| \leq \sqrt{5}$, with equality only for $u=-3$. To prove this, square to get the equivalent $$ (u-1)^2 + (u+1)^2 + 5 - 2\sqrt{(u^2-2u+5)(u^2+2u+2)} \leq 5$$i.e. $u^2 + 1 \leq \sqrt{(u^2-2u+5)(u^2+2u+2)}$. Squaring once more leads to the equivalent $(u+3)^2 \geq 0$, which is clearly true.