From Cauchy-Schwarz inequality we have $$\sqrt{x-1}+\sqrt{2x+9}+\sqrt{19-3x}\le\sqrt{3(x-1+2x+9+19-3x)}=\sqrt{3\cdot 27}=9.$$We would have equality if and only if $x-1=2x+9=19-3x$,which is not possible.Thus the inequality is strict.

Observe that $\frac{\sqrt{x-1}+\sqrt{2x+9}+\sqrt{19-3x}}{3}\leq \sqrt{\frac{(x-1)+(2x+9)+(19-3x)}{3}}=3.$ So $\sqrt{x-1}+\sqrt{2x+9}+\sqrt{19-3x}\leq 9$ with equality iff $\sqrt{x-1}=\sqrt{2x+9}=\sqrt{19-3x}$. This obviosly cannot occur, therefore $\sqrt{x-1}+\sqrt{2x+9}+\sqrt{19-3x}<9$.

It is more beautiful . Let $x$ be real . Prove that $$\sqrt{x-1}+\sqrt{2x-3}+\sqrt{7-3x}\leq 3.$$ See also here China Second Round Olympiad 2003 Q13

Solution By AM-QM $\sqrt{x-1}+\sqrt{2x-3}+\sqrt{7-3x}\leq \sqrt{3(x-1+2x-3+7-3x)}=3$ with equality iff they are all equall $\iff x=1$

WolfusA wrote: Solution By AM-QM $\sqrt{x-1}+\sqrt{2x-3}+\sqrt{7-3x}\leq \sqrt{3(x-1+2x-3+7-3x)}=3$ with equality iff they are all equall $\iff x=1$ maybe equality doesn't hold in this case as the $2x-3$ would be negative?

Sorry, equality iff $x=2$

Cauchy gives us \[(\sqrt{x-1}+\sqrt{2x+9}+\sqrt{19-3x})^2\geq (1+1+1)(x-1+2x+9+19-3x)=81.\]Square rooting finishes.

Alternatively since $f(x) = \sqrt{x}$ is concave Jensen's gives us $\sqrt{x-1} + \sqrt{2x+9} + \sqrt{19-3x} < 3 \sqrt{\frac{x-1+2x+9+19-3x}3} = 9$ with equality iff $x-1 = 2x+9 = 19-3x$, which, as others have said, is impossible.