# Why does $a+b+c=3\;$ imply $\;a\sqrt{a+3} + b\sqrt{b+3} + c\sqrt{c+3} \ge 6$ [duplicate]

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

• Are you familiar with the AMGM inequality theorem that would state that $a+b+c \ge 3\sqrt{abc}$? May 11, 2017 at 18:48
• ... and that $a\sqrt{a+3} + b\sqrt{b+3}+ c\sqrt{b+3} \ge 3\sqrt{abc\sqrt{(a+3)(b+3)(c+3)}}$? May 11, 2017 at 18:50
• 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. May 11, 2017 at 18:53
• Please anyone trying to edit the title, we need words in the title that **do not appear in a hyperlinked/mathjaxed expressions. May 11, 2017 at 18:58
• 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) May 11, 2017 at 19:02

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
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.