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Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that $$a\sqrt{a+3} + b\sqrt{b+3} + c\sqrt{c+3} \geqslant 6$$

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  • $\begingroup$ Are you familiar with the AMGM inequality theorem that would state that $a+b+c \ge 3\sqrt{abc}$? $\endgroup$
    – fleablood
    May 11, 2017 at 18:48
  • $\begingroup$ ... and that $a\sqrt{a+3} + b\sqrt{b+3}+ c\sqrt{b+3} \ge 3\sqrt{abc\sqrt{(a+3)(b+3)(c+3)}}$? $\endgroup$
    – fleablood
    May 11, 2017 at 18:50
  • $\begingroup$ Could you show us your work before having posted your question here? That's kind of the point: we are here to help (advise, hint, suggest, get you started, not to do someone's work for them. $\endgroup$
    – amWhy
    May 11, 2017 at 18:53
  • $\begingroup$ Please anyone trying to edit the title, we need words in the title that **do not appear in a hyperlinked/mathjaxed expressions. $\endgroup$
    – amWhy
    May 11, 2017 at 18:58
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    $\begingroup$ Why are bare problem statements/homework, tagged as inequalities, not addressed like such problems posted in other tags. Lack of context is lack of context, whatever the tag. The asker simply demands us to work on his/her behalf, given some statement, "prove that .... .... " after which, it seems, trigger-happy beavers jump in to oblige/cater to the asker's demand(s) $\endgroup$
    – amWhy
    May 11, 2017 at 19:02

2 Answers 2

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Note that $f(x) = x\sqrt{x +3}$ is convex over $\mathbb{R}^{+}$, hence by Jensen's inequality we have:

$\Rightarrow f\left(\dfrac{a+b+c}{3}\right)\leq \dfrac{f(a) +f(b)+f(c)}{3}$

$\Rightarrow 6 \leq a\sqrt{a+3} +b\sqrt{b+3} +c\sqrt{c+3}$, as required

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    $\begingroup$ Talking about convex functions goes way beyond pre-calculus, according to the OP's tag... $\endgroup$
    – Hans Olo
    May 11, 2017 at 18:55
  • $\begingroup$ @loki123 in all fairness, so does finding the minimum of the function in your answer... $\endgroup$
    – gt6989b
    May 11, 2017 at 18:57
  • $\begingroup$ Actually, the tag "algebra-precalculus" was added by an earlier editor, not by the asker. $\endgroup$
    – amWhy
    May 11, 2017 at 19:10
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First set up a function $$f(a,b,c)=\sqrt{a+3} a+\sqrt{b+3} b+\sqrt{c+3} c.$$ Then, set $a=3-b-c$ to eliminate $a$ and get: $$f(b,c)=-\sqrt{-b-c+6} (b+c-3)+\sqrt{b+3} b+c \sqrt{c+3},$$ then find the minimum of the function $f(b,c)$, which is $f(b=1,c=1)=6$. Therefore, we have $$f \ge 6$$ as requested.

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