By AM-GM $\frac{a+8}{a}+9 \geq 6\sqrt{\frac{a+8}{a}}$ Then $\sqrt{\frac{a+8}{a}} \leq \frac{\frac{a+8}{a}+9}{6}=\frac{5a+4}{3}$ THen $\sqrt{\frac{a}{a+8}} \geq \frac{3}{5a+4}$ The $\sqrt{\frac{a}{a+8}}+\sqrt{\frac{b}{b+8}}+\sqrt{\frac{c}{c+8}} \geq \frac{3}{5a+4}+\frac{3}{5b+4}+\frac{3}{5c+4}$ We put $a=e^x, a=e^y, a=e^z $ then $ \frac{3}{5e^x+4}+\frac{3}{5e^y+4}+\frac{3}{5e^z+4} \geq 3\frac{3}{5e^{x+y+z}+4} = 3\frac{3}{5abc+4}=1$

utkarshgupta wrote: Let $a,b,c \in \mathbb{R^+}$ such that $abc=1$. Prove that $$\sum_{a,b,c} \sqrt{\frac{a}{a+8}} \ge 1$$ By AM-GM,we have \[\sqrt{\frac{a}{a+8}}=\frac{2a(a+2)}{2\sqrt{a(a+8)\cdot(a+2)^2}}\ge{\frac{2a(a+2)}{a(a+8)+(a+2)^2}}=\frac{a(a+2)}{a^2+6a+2}\]Hence,we need to prove that \[\sum{\frac{a(a+2)}{a^2+6a+2}}\ge{1}\]\[<=>\sum{\frac{a(a+2)}{a^2+6a+2}}-1=\frac{\sum{(11b+11c+6bc+cb^2+c^2b+6c^2b^2)(a-1)^2}}{3(a^2+6a+2)(b^2+6b+2)(c^2+6c+2)}\ge{0}\]

Old. Where?

2001,IMO

Substitute $a= \frac {x^2}{yz}$... then apply holder to get the desired inequality

There are so many methods, this inequality is very old

http://www.artofproblemsolving.com/community/c6h17451p1149321

utkarshgupta wrote: Let $a,b,c \in \mathbb{R^+}$ such that $abc=1$. Prove that $$\sum_{a,b,c} \sqrt{\frac{a}{a+8}} \ge 1$$ Let $a,b,c \in \mathbb{R^+}$ such that $abc=1$. Prove that $$\sqrt{\frac{a}{b+3}}+\sqrt{\frac{b}{c+3}}+\sqrt{\frac{c}{a+3}} \ge \frac{3}{2}.$$

sqing wrote: Let $a,b,c \in \mathbb{R^+}$ such that $abc=1$. Prove that $$\sqrt{\frac{a}{b+3}}+\sqrt{\frac{b}{c+3}}+\sqrt{\frac{c}{a+3}} \ge \frac{3}{2}.$$ It's known: http://www.artofproblemsolving.com/community/c6h142431

Thank arqady.

solution by tuchuyenhg.

Attachments:
10.docx (27kb)

i liked it

i think jensen's(product form) will kill the inequality.

ali3985 wrote: By AM-GM $\frac{a+8}{a}+9 \geq 6\sqrt{\frac{a+8}{a}}$ Then $\sqrt{\frac{a+8}{a}} \leq \frac{\frac{a+8}{a}+9}{6}=\frac{5a+4}{3}$ THen $\sqrt{\frac{a}{a+8}} \geq \frac{3}{5a+4}$ The $\sqrt{\frac{a}{a+8}}+\sqrt{\frac{b}{b+8}}+\sqrt{\frac{c}{c+8}} \geq \frac{3}{5a+4}+\frac{3}{5b+4}+\frac{3}{5c+4}$ We put $a=e^x, a=e^y, a=e^z $ then $ \frac{3}{5e^x+4}+\frac{3}{5e^y+4}+\frac{3}{5e^z+4} \geq 3\frac{3}{5e^{x+y+z}+4} = 3\frac{3}{5abc+4}=1$ The rhs of second line would be $\frac{5a+4}{3a}$, so that the function on which you apply Jensen's inequality, should be $ \frac{3e^x}{5e^x+4} .$ unfortunately, this is NOT convex for every $x.$ And yes, as some already said before, that after a little substitution, $a=\frac{x^2}{yz}$ etc., It becomes nothing but the well-known $\frac{a}{\sqrt{a^2+8bc}}$ (IMO 2001) problem.