Prove: Inequality $9\gt \sqrt{a-1}+\sqrt{19-3a}+\sqrt{2a+9}$

[anemone]

Prove that $9\gt \sqrt{a-1}+\sqrt{19-3a}+\sqrt{2a+9}$ for all real $a$.

 

[Euge]

Spoiler

The inequality only makes sense when $1\le a\le 19/3$. By concavity of the square root function on $(0,\infty)$,

$$\sqrt{a-1}+\sqrt{19-3a}+\sqrt{2a+9} < 3\sqrt{\frac{(a-1)+(19-3a)+(2a+9)}{3}} = 3\sqrt{\frac{27}{3}} = 3\cdot 3 = 9.$$

 

[anemone]

Spoiler

The inequality only makes sense when $1\le a\le 19/3$. By concavity of the square root function on $(0,\infty)$,

$$\sqrt{a-1}+\sqrt{19-3a}+\sqrt{2a+9} < 3\sqrt{\frac{(a-1)+(19-3a)+(2a+9)}{3}} = 3\sqrt{\frac{27}{3}} = 3\cdot 3 = 9.$$

Thanks Euge for your elegant method of proving and thanks too for participating.