Calculation of $\bf{Max.}$ value of $\sqrt{10x-x^2}-\sqrt{18x-x^2-77}\;\;\forall x\in \mathbb{R}$ (1) Calculation of Max. and Min. value of $\sqrt{x^2-3x+2}+\sqrt{2+3x-x^2}\;\; \forall x\in \mathbb{R}$
(2) Calculation of Max. value of $\sqrt{10x-x^2}-\sqrt{18x-x^2-77}\;\;\forall x\in \mathbb{R}$
My Try:: for $(1)$ one:: Using the Cauchy-Schwarz inequality
$$\Rightarrow \displaystyle \left\{\left(\sqrt{x^2-3x+2}\right)^2+\left(\sqrt{2+3x-x^2}\right)^2\right\}\cdot \left(1^2+1^2\right)\geq \left\{\sqrt{x^2-3x+2}+\sqrt{2+3x-x^2}\right\}^2$$
$$\Rightarrow \displaystyle \left\{\sqrt{x^2-3x+2}+\sqrt{2+3x-x^2}\right\}^2\leq 8$$
$$\Rightarrow \left\{\sqrt{x^2-3x+2}+\sqrt{2+3x-x^2}\right\}\leq 2\sqrt{2}$$
and equality hold when $\displaystyle \sqrt{x^2-3x+2}=\sqrt{2+3x-x^2}\Rightarrow x=0\;,3$
Is my solution right for Max.?  If not then please help me, and how can I calculate for Min. and also help required in $(2)$ one.

(1):the approach is not right ,you can't make sure $a\ge b,b\le c \to a \ge c $
the easy way is let$ u=x^2-3x, f(x)=f(u)=\sqrt{2+u}+\sqrt{2-u},f^2=4+2\sqrt{4-u^2}$
$0\le\sqrt{4-u^2} \le 2 \implies 4 \le f^2 \le 8 \implies 2 \le f \le 2\sqrt{2}$  when $u=0$ and $u=2$ get Max and Min.
(2): $\sqrt{10x-x^2}$ domain is [$0,10$], $\sqrt{18x-x^2-77}$ domain is [$7,11$] so  $f(x)=\sqrt{10x-x^2}-\sqrt{18x-x^2-77}$ domain is [$7,10$]
$\sqrt{10x-x^2}$ is mono decreasing on [$7,10$], $\sqrt{18x-x^2-77}$ is increasing on [$7,9$] so we can sure $f(x)$ will get Max when $x=7$ when $x$ on [$7,9$], the remain is [$9,10$],but it is impossible for any $f(x) \ge f(7)$ because it is always $ <f(7)- a, a >0$ 
