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

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The inequality \(9 > \sqrt{a-1} + \sqrt{19-3a} + \sqrt{2a+9}\) holds true for all real values of \(a\). The discussion highlights an elegant proof method provided by a user named Euge, which effectively demonstrates the validity of the inequality across the entire set of real numbers. The proof utilizes properties of square roots and inequalities to establish the conclusion definitively.

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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 $(0,\infty)$,

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