Problem 1512

To prove that $\sqrt{x-1}+\sqrt{2x+9}+\sqrt{19-3x}<9$ for all real $x$ for which the left-hand side is well defined, we will use the Arithmetic Mean - Quadratic Mean (AM-QM) inequality.

1. Determine the domain of the function:
The expression $\sqrt{x-1}+\sqrt{2x+9}+\sqrt{19-3x}$ is well defined if and only if all the expressions under the square roots are non-negative:
$$x-1 \geq 0 \implies x \geq 1,$$
$$2x+9 \geq 0 \implies x \geq -\frac{9}{2},$$
$$19-3x \geq 0 \implies x \leq \frac{19}{3}.$$
Combining these conditions, the domain is:
$$1 \leq x \leq \frac{19}{3}.$$

2. Apply the AM-QM inequality:
The AM-QM inequality states that for any non-negative real numbers $a, b, c$:
$$\frac{a+b+c}{3} \leq \sqrt{\frac{a^2+b^2+c^2}{3}}.$$
Let $a = \sqrt{x-1}$, $b = \sqrt{2x+9}$, and $c = \sqrt{19-3x}$. Then:
$$\frac{\sqrt{x-1}+\sqrt{2x+9}+\sqrt{19-3x}}{3} \leq \sqrt{\frac{(\sqrt{x-1})^2 + (\sqrt{2x+9})^2 + (\sqrt{19-3x})^2}{3}}.$$

3. Simplify the right-hand side:
$$\sqrt{\frac{(x-1) + (2x+9) + (19-3x)}{3}} = \sqrt{\frac{x-1 + 2x+9 + 19-3x}{3}} = \sqrt{\frac{27}{3}} = \sqrt{9} = 3.$$
Therefore:
$$\frac{\sqrt{x-1}+\sqrt{2x+9}+\sqrt{19-3x}}{3} \leq 3.$$

4. Multiply both sides by 3:
$$\sqrt{x-1}+\sqrt{2x+9}+\sqrt{19-3x} \leq 9.$$

5. Check for equality:
Equality holds in the AM-QM inequality if and only if $\sqrt{x-1} = \sqrt{2x+9} = \sqrt{19-3x}$. Squaring these equalities, we get:
$$x-1 = 2x+9 = 19-3x.$$
Solving $x-1 = 2x+9$:
$$x-1 = 2x+9 \implies -x = 10 \implies x = -10,$$
which is not in the domain $1 \leq x \leq \frac{19}{3}$.

Solving $2x+9 = 19-3x$:
$$2x+9 = 19-3x \implies 5x = 10 \implies x = 2,$$
which is in the domain. However, substituting $x = 2$ into the original expressions:
$$\sqrt{2-1} = 1, \quad \sqrt{2(2)+9} = \sqrt{13}, \quad \sqrt{19-3(2)} = \sqrt{13},$$
we see that $1 + \sqrt{13} + \sqrt{13} = 1 + 2\sqrt{13}$, which is not equal to 9. Therefore, equality cannot hold.

6. Conclusion:
Since equality cannot hold, we have:
$$\sqrt{x-1}+\sqrt{2x+9}+\sqrt{19-3x} < 9.$$

The final answer is $\boxed{ \sqrt{x-1}+\sqrt{2x+9}+\sqrt{19-3x} < 9 }$.