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Let $a,b,c$ be reals satisfying:

(i) $a,b,c\ge0$

(ii) $a+b+c=4$

Find the minimum value of the expression

$\sqrt {2a+1}$ $+$ $\sqrt {2b+1}$ $+$ $\sqrt {2c+1}$

So I am literally clueless - I know that the maximum value of the expression can be found by using QM-AM (because the expression is less than or equal to the square root of 3 times the sum of the squares of each term) but I don't know how to find the minimum.

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  • $\begingroup$ Hint: That is a concave function on a convex domain ... $\endgroup$ – Martin R May 9 '18 at 14:05
  • $\begingroup$ Compare math.stackexchange.com/questions/1693748/… for a similar question. $\endgroup$ – Martin R May 9 '18 at 14:09
  • $\begingroup$ Do you know about Lagrange multipliers? $\endgroup$ – user438666 May 9 '18 at 14:57
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Let $f(x)=\sqrt{2x+1}.$

Hence, $f$ is a concave function.

Thus, by Karamata $$\sum_{cyc}\sqrt{2a+1}\geq f(a+b+c)+2f(0)=3+2\cdot1=5.$$

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Hint: As $ t \mapsto \sqrt t$ is concave, use Karamata’s inequality with $(a+b+c,0,0) \succeq (a,b,c)$.

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