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

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Prove that $9\gt \sqrt{a-1}+\sqrt{19-3a}+\sqrt{2a+9}$ for all real $a$.
 
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The inequality only makes sense when $1\le a\le 19/3$. By concavity of the square root function on $(0,\infty)$,

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