$\frac{b+c}{a}+2\ge2\sqrt{\frac{2(b+c)}{a}}=2\sqrt{2}\sqrt{\frac{1-a}{a}}$

Let $t = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$ From the Cauchy-Schwarz inequality, we have \[\sqrt{2t} = \sqrt{2\cdot \sum_{cyc} \frac{1}{a}} \ge \sum_{cyc} \sqrt{\frac{1-a}{a}}.\] Notice that $6 + \sum_{sym} \frac{a}{b} = 3 + t$, so it remains to prove that $3 + t \ge 4\sqrt{t}$, which follows because $t \ge 9$ and \[ 9 + t - 6 \ge 6\sqrt{t} - 6 \ge 4 \sqrt{t}.\]

We have $\displaystyle \frac {a} {b} + \frac {b} {a} + \frac {b} {c} + \frac {c} {b} + \frac {c} {a} + \frac {a} {c} = \sum \frac {b+c} {a} = \sum \frac {1-a} {a}$, so \[\frac {a} {b} + \frac {b} {a} + \frac {b} {c} + \frac {c} {b} + \frac {c} {a} + \frac {a} {c} + 6 - 2\sqrt{2} (\sqrt{\frac {1-a} {a}} + \sqrt{\frac {1-b} {b}} + \sqrt{\frac {1-c} {c}} ) =\] \[{\sum ( \frac {1-a} {a}} - 2\sqrt{2}\sqrt{\frac {1-a} {a}} + 2 ) = \sum ( \sqrt{\frac {1-a} {a}} - \sqrt{2} )^2 \geq 0.\] Equality occurs when $\displaystyle \sqrt{\frac {1-a} {a}} = \sqrt{\frac {1-b} {b}} = \sqrt{\frac {1-c} {c}} = \sqrt{2}$, thus for $a=b=c=\dfrac {1} {3}$.

Generalization a)Let $x,y,z\ge 0$ and $a,b,c,\alpha ,\beta ,\gamma $ be positive real numbers such that $a+b+c=1$. Prove that \[\beta \frac{a}{b+y}+\gamma \frac{a}{c+z}+\beta \frac{c}{b+y}+\alpha \frac{c}{a+x}+\gamma \frac{b}{c+z}+\alpha \frac{b}{a+x}+2\left( \frac{\alpha }{1+3x}+\frac{\beta }{1+3y}+\frac{\gamma }{1+3z} \right)\ge \] \[2\sqrt{2}\left( \sqrt{\frac{{{\alpha }^{2}}}{1+3x}\cdot \frac{1-a}{x+a}}+\sqrt{\frac{{{\beta }^{2}}}{1+3y}\cdot \frac{1-b}{y+b}}+\sqrt{\frac{{{\gamma }^{2}}}{1+3z}\cdot \frac{1-c}{z+c}} \right).\] When does equality hold? b)Let and $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that \[\frac{a}{b+1}+\frac{a}{c+1}+\frac{c}{b+1}+\frac{c}{a+1}+\frac{b}{c+1}+\frac{b}{a+1}+\frac{3}{2}\ge \] \[\sqrt{2}\left( \sqrt{\frac{1-a}{1+a}}+\sqrt{\frac{1-b}{1+b}}+\sqrt{\frac{1-c}{1+c}} \right).\] When does equality hold?

$y=\sqrt{x}\ (x>0)$ is an increasing and concave function, thus by Jensen's theorem, we have \[{\frac{\sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}}{3}\leq \sqrt{\frac{\frac{1-a}{a}+\frac{1-b}{b}+\frac{1-c}{c}}{3}}}\] Since $0<a,\ b,\ c<1$, using the $A.M.-G.M.$ inequality, \[2\sqrt{2}\left(\sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}\right)\leq 2\sqrt{6}\sqrt{\frac{1-a}{a}+\frac{1-b}{b}+\frac{1-c}{c}}\leq 6+\frac{1-a}{a}+\frac{1-b}{b}+\frac{1-c}{c}=\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}+6.\] The equality holds when $\frac{1-a}{a}=\frac{1-b}{b}=\frac{1-c}{c}$ and $a+b+c=1\Longleftrightarrow a=b=c=\frac 13$, which satisfy $0<a,\ b,\ c<1.$

Who knows the official results of this contest? Please post it or post jbmo 2012 official site

There is this topic http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&t=485894&p=2732757#p2732757 which gave the link to the site in due time. For convenience, here it is http://www.hms.gr/16jbmo2012/main.htm.

Let $x=\frac{a+b}{c}, y=\frac{b+c}{a}, z=\frac{c+a}{b}$, clearly all positive reals. Then, $(x+y+z-6)^2 \geq 0 \implies (x+y+z)^2-12(x+y+z)+36 \geq 0$ $\implies (x+y+z)^2+12(x+y+z)+36 \geq 24(x+y+z)$ $\implies (x+y+z+6)^2 \geq 24(x+y+z) = 8(x+y+z)(1+1+1) \geq 8(\sqrt{x}+\sqrt{y}+\sqrt{z})^2$ (Cauchy) $\implies x+y+z+6 \geq 2\sqrt{2}(\sqrt{x}+\sqrt{y}+\sqrt{z})^2$ $\implies \frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+6 \geq 2\sqrt{2}(\frac{1-a}{a}+\frac{1-b}{b}+\frac{1-c}{c})$. Where the last step follows from $a+b+c=1$. Equality occurs when $\frac{x}{1}=\frac{y}{1}=\frac{z}{1} \implies x=y=z \implies a=b=c=\frac{1}{3}$.

Using the $a+b+c=1$ constraint , the LHS rewrites as $ \sum {\frac{1-a}{a}} +6 $ And we have $\frac{1-a}{a}+2 \geq 2\sqrt{2}\sqrt{\frac{1-a}{a}}$ Do the same for others and sum up to the desired inequality.

Solved with AM-GM

using $a+b+c=1$ in LHS , the inequality is equivalent to $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} +3 \geq 2\sqrt{2}\left (\sqrt{\frac{1-a}{a}} + \sqrt{\frac{1-b}{b}} + \sqrt{\frac{1-c}{c}}\right ). $ use Cauchy, $3(\frac{1}{a}+\frac{1}{b}+\frac{1}{c} -3) \geq \left(\sqrt{\frac{1-a}{a}} + \sqrt{\frac{1-b}{b}} + \sqrt{\frac{1-c}{c}}\right )^2. $ so it remains to prove $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c} +3)^2 \geq 24 \left (\frac{1}{a}+\frac{1}{b}+\frac{1}{c} -3\right )$ let $p=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ , we have $(p-9)^2 \geq 0 . $

I used a combination of AM-GM and Cauchy

I used a combination of AM-GM and Cauchy

To be honest, I think the most simple solution is to go with a / b + c / b = ( a + c ) / b = ( 1 - b ) / b, 2 + ( 1 - b ) / b = > 2 * sqrt [ ( 1 - b ) / b ] (by AM-GM), and repeating the process for the other ones. No Cauchy and no Lemmas so it is very simple yet very quick solution, easy to explain to beginners like me

guiguzi wrote: $\frac{b+c}{a}+2\ge2\sqrt{\frac{2(b+c)}{a}}=2\sqrt{2}\sqrt{\frac{1-a}{a}}$ Nice solution

For storage. $$\frac {a}{b} + \frac {a}{c} + \frac {c}{b} + \frac {c}{a} + \frac {b}{c} + \frac {b}{a} + 6 \geq 2\sqrt{2}\left (\sqrt{\frac{1-a}{a}} + \sqrt{\frac{1-b}{b}} + \sqrt{\frac{1-c}{c}}\right )$$ $$\iff \left(\sum \frac{a+b}{c}\right) + 6 \ge 2\sqrt{2} \left( \sum \sqrt{\frac{1 - c}{c}} \right)$$ $$\iff \sum \left( \frac{a+b}{c} + 2\right) \ge \sum 2\sqrt{\frac{2(a+b)}{c}}$$ $$\iff \sum \left( \sqrt{\frac{a+b}{c}} - \sqrt{2}\right)^2 \ge 0$$Which is the trivial inequality. Equality holds when $\boxed{a = b = c = \frac 13}$

Notice:\begin{align*} &\frac{1-x}{x}-2\sqrt{\frac{2(1-x)}{x}} \geq -2 \\& \iff \frac{1+x}{x}\geq \frac{2\sqrt{2(1-x)}}{\sqrt{x}} \\& \iff (3x-1)^2\geq 0.\end{align*}By applying this to the original inequality, the claim follows. Note that equality when $a=b=c=\frac{1}{3}.$

emregirgin35 wrote: Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that \[\frac {a}{b} + \frac {a}{c} + \frac {c}{b} + \frac {c}{a} + \frac {b}{c} + \frac {b}{a} + 6 \geq 2\sqrt{2}\left (\sqrt{\frac{1-a}{a}} + \sqrt{\frac{1-b}{b}} + \sqrt{\frac{1-c}{c}}\right ).\]When does equality hold? $$\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}+6\ge 2\sqrt2\left(\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}}\right)$$$$\iff$$$$\left(\sqrt{\frac{b+c}{a}}-\sqrt 2\right)^2+\left(\sqrt{\frac{c+a}{b}}-\sqrt 2\right)^2+\left(\sqrt{\frac{a+b}{c}}-\sqrt 2\right)^2\geq 0$$2019 Saudi Arabia Kosovo National MO here here

probably overcomplicated... we need to prove $(3+(a+b+c)(\frac{1}{a} + \frac{1}{b} + \frac{1}{b})^2 \ge 8(\sqrt{\frac{1-a}{a}} + \sqrt{\frac{1-b}{b}} + \sqrt{\frac{1-b}{b}})^2$. Note that $(\sqrt{\frac{1-a}{a}} + \sqrt{\frac{1-b}{b}} + \sqrt{\frac{1-b}{b}})^2 \le ((1-a)+(1-b)+(1-c))(\frac{1}{a} + \frac{1}{b} + \frac{1}{b})$ so we need to prove $(3+(\frac{1}{a} + \frac{1}{b} + \frac{1}{b}))^2 \ge 16(\frac{1}{a} + \frac{1}{b} + \frac{1}{b})$. Note that $(a+b+c)(\frac{1}{a} + \frac{1}{b} + \frac{1}{b}) \ge (1+1+1)^2 \implies (\frac{1}{a} + \frac{1}{b} + \frac{1}{b}) \ge 9$ so Let $(\frac{1}{a} + \frac{1}{b} + \frac{1}{b}) = 9 + t$ now we need to prove $(12+t)^2 \ge 16(9+t)$ which is true and equality holds for $t = 0$.

Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that $$\frac {a}{b} + \frac {a}{c} + \frac {c}{b} + \frac {c}{a} + \frac {b}{c} + \frac {b}{a} + 6 \geq\sqrt{6}\left (\sqrt{\frac{1-a^2}{a}} + \sqrt{\frac{1-b^2}{b}} + \sqrt{\frac{1-c^2}{c}}\right )$$

$$\text{First part:}$$$$\sum{\frac {a+c}{b}}+6=\sum{\frac{1}{a}}+3$$$$\text{Second part:}$$$$\sum{\sqrt{\frac{6-6a^2}{a}}}$$$$\boxed{A.M-G.M}$$The right hand side $$\iff$$$$\sum{\frac{1}{2a}}+\frac{15}{2}$$And we get by some analysis $$\sum{\frac{1}{a}} \geq 9$$Which is true for $$\boxed{A.M-H.M}$$So we are done...

sqing wrote: Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that $$\frac {a}{b} + \frac {a}{c} + \frac {c}{b} + \frac {c}{a} + \frac {b}{c} + \frac {b}{a} + 6 \geq\sqrt{6}\left (\sqrt{\frac{1-a^2}{a}} + \sqrt{\frac{1-b^2}{b}} + \sqrt{\frac{1-c^2}{c}}\right )$$ My sol))

sqing wrote: Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that $$\frac {a}{b} + \frac {a}{c} + \frac {c}{b} + \frac {c}{a} + \frac {b}{c} + \frac {b}{a} + 6 \geq\sqrt{6}\left (\sqrt{\frac{1-a^2}{a}} + \sqrt{\frac{1-b^2}{b}} + \sqrt{\frac{1-c^2}{c}}\right )$$ By $AM-GM$, we have $$\frac{b+c}{a}+\frac{3}{2}(a+1) \ge 2\sqrt{\frac{3}{2} \cdot \frac{(a+1)(b+c)}{a}}$$$$\Rightarrow \frac{b+c}{a}+ \frac{3}{2}(a+1) \ge \sqrt{6} \sqrt{\frac{(a+1)(1-a)}{a}}= \sqrt{6} \sqrt{\frac{1-a^2}{a}}$$Summing it cyclically, we get the desired inequality.

AM-GM and C-S gives the desired result.