Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that $$a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3} \geq 6.$$
Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam
Solution:
Let $f(x)=x\sqrt{x+3}$. Observe that $$f''(x)=\dfrac{3(x+4)}{4\sqrt{(x+3)^3}}>0 \qquad \forall x>0.$$ Therefore, by Jensen's Inequality, we have $$2=f(1)=f\left(\dfrac{a+b+c}{3}\right) \leq \dfrac{f(a)+f(b)+f(c)}{3}=\dfrac{a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3}}{3},$$ i.e. $$a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3} \geq 6.$$
Let $f(x)=x\sqrt{x+3}$. Observe that $$f''(x)=\dfrac{3(x+4)}{4\sqrt{(x+3)^3}}>0 \qquad \forall x>0.$$ Therefore, by Jensen's Inequality, we have $$2=f(1)=f\left(\dfrac{a+b+c}{3}\right) \leq \dfrac{f(a)+f(b)+f(c)}{3}=\dfrac{a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3}}{3},$$ i.e. $$a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3} \geq 6.$$
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