$ \frac 1{1+a^2(b+c)} + \frac 1{1+b^2(c+a)} + \frac 1 {1+c^2(a+b)} \leq \frac 3 {1+2abc}$ $ \iff \frac 1{1+a^2(b+c)}-\frac{1}{1+2abc} +\frac 1{1+b^2(c+a)}-\frac{1}{1+2abc} + \frac 1 {1+c^2(a+b)}-\frac{1}{1+2abc} \leq 0$ Note that : $ \frac 1{1+a^2(b+c)}- \frac{1}{1+2abc}=\frac{ab(c-a)+ac(b-a)}{1+a^2(b+c)(1+2abc)}$ Hence $ LHS= \frac{abc-1}{2abc+1}[(c-a)^2S_b+(a-b)^2S_c+(b-c)^2S_a)$ where :$ S_b=\frac{b}{(1+a^2(b+c))(1+c^2(a+b))}$ and similar $ S_a=\frac{a}{(1+b^2(a+c))(1+c^2(a+b))}$ and $ S_c=\frac{c}{(1+a^2(b+c))(1+b^2(a+c))}$ with $ S_a,S_b,S_c \geq 0$(*) By ineq AM-GM : $ ab+bc+ca \geq 3\sqrt[3]{a^2b^2c^2} \iff abc \leq 1$(**) From (*) and (**), we have : $ LHS \leq 0$ Occur equal if and only if :$ a=b=c=1$

From $ ab + bc + ca = 3$(multiply by $ a$) we have: $ a^2(b + c) = 3a - abc$ , so it's enough to prove the following inequality: \[ \sum\frac {1}{1 - abc + 3a} \leq \frac {3}{1 + 2abc} \] Let $ p = 1 - abc$ , obviously $ p \geq 0$ because with AM-GM inequality we have: $ 3 = ab + bc + ca \geq 3\sqrt [3]{a^2b^2c^2}$ so we have $ abc \leq 1$ , with equality if and only if $ a = b = c$. With our new marks, we have to prove the following inequality: \[ \sum\frac {1}{p + 3a} \leq \frac {3}{3 - 2p} \] or, after expanding: \[ (3 - 2p)(p^2 + 2p(a + b + c) + 27) \leq (p + 3a)(p + 3b)(p + 3c) \] \[ \leftrightarrow 3p^2 + 6p(a + b + c) + 27(1 - abc) - 2p^3 - 4p^2(a + b + c) - 18p \leq p^3 + p^2(3a + 3b + 3c) + 27p \] \[ \leftrightarrow p(3p^2 + p(7a + 7b + 7c) + 18 - 6a - 6b - 6c) \geq 0 \] Since $ p \geq 0$ , it's enough to prove the thing inside the brackets is $ \geq 0$. Now if we return $ p = 1 - abc$ and expand: \[ p(3a^2b^2c^2 + 3 - 6abc + 18 + a + b + c - 7abc( a + b + c )) \geq 0 \] $ 3a^2b^2c^2 + 3 \geq 6abc$ is due to AM-GM; $ 18 \geq 6abc(a+b+c)$ is the same as : $ T[2,2,0]+2T[2,1,1] \geq 3T[2,1,1]$ which is Muirhead , and: $ a+b+c \geq abc(a+b+c)$ because $ abc \leq 1$ Summing up last inequalities we get the wanted one. Equality if and only if $ a=b=c$.

My solution is the following The inequality is equivalent to \[ \frac{a^2(b+c)}{1+a^2(b+c)} +\frac{b^2(c+a)}{1+b^2(c+a)} +\frac{c^2(a+b)}{1+c^2(a+b)} +\frac{3}{1+2abc} \ge 3\] By the Cauchy Schwarz Inequality, we get \[ \frac{a^2(b+c)}{1+a^2(b+c)} +\frac{b^2(c+a)}{1+b^2(c+a)} +\frac{c^2(a+b)}{1+c^2(a+b)} \ge \frac{36}{(b+c)[1+a^2(b+c)]+(c+a)[1+b^2(c+a)]+(a+b)[1+c^2(a+b)]} =\frac{18}{(ab+bc+ca)^2 +(a+b+c)(1-abc)} =\frac{18}{9+(a+b+c)(1-abc)}\] Hence, it suffices to prove that \[ \frac{6}{9+(a+b+c)(1-abc)} +\frac{1}{2abc+1} \ge 1\] Setting $ p=a+b+c,r=abc$, then the above inequality is equivalent to \[ \frac{6}{9+p(1-r)} +\frac{1}{2r+1} \ge 1\] \[ \Leftrightarrow \frac{6}{9+p(1-r)} \ge \frac{2r}{1+2r}\] \[ \Leftrightarrow 3(2r+1) \ge r[9+p(1-r)]\] \[ \Leftrightarrow (1-r)(3-pr) \ge 0\] Which is obviously true since \[ 27=(ab+bc+ca)^3 \ge 27a^2b^2c^2=27r^2 \Rightarrow 1 \ge r\] \[ 9=(ab+bc+ca)^2 \ge 3abc(a+b+c)=3pr \Rightarrow 3 \ge pr\] We have done. Equality holds if and only if $ a=b=c=1$. @Moderators: How can I propose a problem to the contest? Please reply me as soon as possible. Thank you very much.

We have $ \frac{1}{{1 + a^2 \left( {b + c} \right)}} = 1 - \frac{{a^2 \left( {b + c} \right)}}{{1 + a^2 \left( {b + c} \right)}}$ So by Cauchy Schwarz we have \[ \sum {\frac{1}{{1 + a^2 \left( {b + c} \right)}}} = 3 - \sum {\frac{{a^2 \left( {b + c} \right)}}{{1 + a^2 \left( {b + c} \right)}}} \le 3 - \frac{{4\left( {ab + bc + ca} \right)^2 }}{{2\left( {a + b + c} \right) + 2\sum {a^2 b^2 } + 2abc\left( {a + b + c} \right)}}\] Assume $ p=a+b+c,r=abc$,it suffices us to show that \[ 3 - \frac{{36}}{{2p + 18 - 2rp}} \le \frac{3}{{1 + 2r}}\] \[ \Leftrightarrow pr \le 3\] which is clearly true because we have $ \left( {ab + bc + ca} \right)^2 \ge 3abc\left( {a + b + c} \right) \Leftrightarrow pr \le 3$. We have done, equality holds if and only if $ a=b=c=1$.

We have to prove that $ \sum\frac {1}{1 + a^2(b + c)}\leq\frac {3}{1 + 2abc}$ or equivalently $ \sum\left(\frac {1}{1 + a^2(b + c)} - \frac {1}{1 + 2abc}\right)\leq 0$ or $ \sum\left(\frac {a(b(a - c) + c(a - b))}{(1 + a^2(b + c))(1 + 2abc)}\right)\geq 0$ or $ \sum\left(\frac {ac(a - b)}{(1 + a^2(b + c))(1 + 2abc)} + \frac {bc(b - a)}{(1 + b^2(a + c))(1 + 2abc)}\right)\geq 0$ or $ \sum\left(\frac {c(a - b)^2(1 - abc)}{(1 + a^2(b + c))(1 + b^2(c + a))(1 + 2abc)}\right)\geq 0$ which holds since $ ab + bc + ca\geq 3\sqrt [3]{a^2b^2c^2}$ which means that $ abc\leq 1$ and we are done .

$ RHS-LHS=\sum_{cyc}(\frac{1}{1+2abc}-\frac{1}{1+a^{2}(b+c)})=\sum_{cyc}\frac{ab(a-c)+ac(a-b)}{(1+2abc)(1+a^{2}(b+c))}=\sum_{cyc}\frac{a(b-c)}{1+2abc}(\frac{b}{1+b^{2}(c+a)}-\frac{c}{1+c^{2}(a+b)})=\sum_{cyc}\frac{a(1-abc)(b-c)^{2}}{(1+2abc)(1+b^{2}(c+a))(1+c^{2}(a+b))}$. Because $ ab+bc+ca=3$ , $ abc\le1$. (AM-GM) Therefore $ RHS\ge LHS$. Equality occurs iff $ a=b=c=1$.

$ \sum\frac {1}{1 + a^2(b + c)}\le \frac {3}{1 + 2abc}\Leftrightarrow \sum(1 - \frac {1}{1 + a^2(b + c)})\ge 3 - \frac {3}{1 + 2abc}\Leftrightarrow \sum\frac {a^2(b + c)}{1 + a^2(b + c)}\ge \frac {6abc}{1 + 2abc}$ By Titu inequality(or cauchy or T'2 Lemma)we have $ \sum\frac {a^2(b + c)}{1 + a^2(b + c)} = \sum\frac {a^2(b + c)^2}{b + c + a^2(b + c)^2}\ge \frac {(2(ab + bc + ca))^2}{2(\sum a) + \sum(ab + ac)^2}$ by the hypothesis we must prove that \[ \frac {36}{2(\sum a) + \sum(3 - bc)^2}\ge \frac {6abc}{1 + 2abc}\Leftrightarrow \frac {6}{2(\sum a) + (\sum (bc)^2) - 6(\sum bc) + 27}\ge \frac {abc}{1 + 2abc} \] Or \[ \frac {6}{2(\sum a) + (\sum (bc)^2) + 9}\ge \frac {abc}{1 + 2abc} \] Now let $ abc = r$ and $ ab + bc + ca = q = 3$ and $ a + b + c = p$ so the last inequality is equivalent to $ \frac {6}{2p + (q^2 - 2pr) + 9}\ge \frac {r}{1 + 2r} \Leftrightarrow 6(1 + 2r)\ge r(2p - 2pr + 18)\Leftrightarrow pr^2 - r(p + 3) + 3\ge 0 \Leftrightarrow p(r - 1)(r - \frac {3}{p})\ge 0 \Leftrightarrow (r - 1)(r - \frac {3}{p})\ge 0$ this inequality is true if we have $ r\le Min(1,\frac {3}{p})$ but by two simple inequality we have $ q^2\ge 3pr$ so $ 3\ge pr$ and $ p^2\ge 3q$ so $ p\ge 3$ So $ \frac {3}{p}\le 1$ $ r\le \frac {3}{p}$ so the inequality is proved and the quality hols for $ a = b = c = 1$

- $ \frac {1}{1 + c^2(a + b)} + \frac {1}{1 + b^2(c + a)} + \frac {1}{1 + a^2(b + c)}\leq \frac {3}{1 + 2abc}$ - denote $ f(x) = \frac {x}{x + k}$ - function $ f(x)$ is concave for all positive real $ x$ and for a positive real parameter $ k$ because $ 0 > f''(x) = \frac { - 2k}{(x + k)^3}$ - if $ ab + bc + ca = 3$ than by AMGM $ 1\geq abc$. if $ abc = 1$ than $ a = b = c = 1$ and inequality holds for this case. so assume without a loss of generality that $ 1 > abc$ - since $ \frac {ab + bc + ca}{3} = 1$ we know that $ \frac {1}{1 + c^2(a + b)} + \frac {1}{1 + b^2(c + a)} + \frac {1}{1 + a^2(b + c)} = \frac {c(a + b) + ab}{3(1 + c^2(a + b))} + \frac {b(c + a) + ca}{3(1 + b^2(c + a))} + \frac {a(b + c) + bc}{3(1 + a^2(b + c))}$ - now by multiplying both sides of the inequality by $ abc$ we get equivalent $ \frac {ab}{3}\frac {c^2(a + b) + abc}{1 + c^2(a + b)} + \frac {ca}{3}\frac {b^2(c + a) + abc}{1 + b^2(c + a)} + \frac {ab}{3}\frac {c^2(a + b) + abc}{1 + c^2(a + b)}\leq\frac {3abc}{1 + 2abc}$ - denote $ k = 1 - abc$. we know $ k > 0$ - left hand side is therefore equivalent to: $ \frac {ab}{3}\frac {c^2(a + b) + abc}{abc + c^2(a + b) + k} + \frac {ca}{3}\frac {b^2(c + a) + abc}{abc + b^2(c + a) + k} + \frac {ab}{3}\frac {c^2(a + b) + abc}{abc + c^2(a + b) + k}$ - denote $ t_1 = \frac {ab}{3}$,,$ t_2 = \frac {ca}{3}$,,$ t_3 = \frac {bc}{3}$. note that $ t_1 + t_2 + t_3 = 1$ - denote $ x = c^2(a + b) + abc$,,$ y = b^2(c + a) + abc$,,$ z = a^2(b + c) + abc$ - left hand side is further equivalent to $ t_1 \frac {x}{x + k} + t_2 \frac {y}{y + k} + t_3 \frac {z}{z + k}$ further equivalent to $ t_1f(x) + t_2f(y) + t_3f(z)$ which by Jensen inequality for concave functions $ t_1f(x) + t_2f(y) + t_3f(z)\leq f(t_1x + t_2y + t_3z)$ since $ x,y,z > 0$ and $ f$ is a concave function for positive reals - $ f(t_1x + t_2y + t_3z) = \frac {\frac {ab}{3}\cdot(c^2(a + b) + abc) + \frac {ca}{3}\cdot(b^2(a + c) + abc) + \frac {bc}{3}\cdot(a^2(b + c) + abc)}{\frac {ab}{3}\cdot(c^2(a + b) + abc) + \frac {ca}{3}\cdot(b^2(a + c) + abc) + \frac {bc}{3}\cdot(a^2(b + c) + abc) + k}$ - since $ c^2(a + b) + abc = c(ab + bc + ca) = 3c$ and analogously for $ b,a$ we get that $ f(t_1x + t_2y + t_3z) = \frac {3abc}{3abc + k} = \frac {3abc}{1 + 2abc}$ $ \frac {ab}{3}\frac {c^2(a + b) + abc}{1 + c^2(a + b} + \frac {ca}{3}\frac {b^2(c + a) + abc}{1 + b^2(c + a)} + \frac {ab}{3}\frac {c^2(a + b) + abc}{1 + c^2(a + b)}\leq\frac {3abc}{1 + 2abc}$ now obviously holds for all positive real $ a,b,c$ such that $ ab + bc + ca = 3$. therefore also the equivalent inequality must hold $ \frac {1}{1 + c^2(a + b)} + \frac {1}{1 + b^2(c + a)} + \frac {1}{1 + a^2(b + c)}\leq \frac {3}{1 + 2abc}$ Q.E.D

Lest find (a+b) from the ab+bc+ac=3 : c(a+b)=3-ab a+b=(3-ab)/c Lets do it for a+c and b+c and put it into our inequality 1/(1+a * (3-bc)) + 1/(1+b * (3-ac)) + 1/(1+c * (3-ba)) <= 3/2+abc 1/(1+3a-abc) + 1/(1+3b-abc)) + 1/(1+3c-abc)) <= 3/2+abc Lets put abc=x 1/(1+3a-x) + 1/(1+3b-x)) + 1/(1+3c-x)) <= 3/2+x 9+2(a+b+c)(1-x)+(1-x)^2 / 9x + 9(1-x)+(a+b+c)(1-x)^2+(1-x)^3 <= 3/2+x After entreating this inequality we will achive : 4x*x*x-9*x*x+15*x-10 <= (a+b+c)(5x*x-4*x-1) If we do the Cauchi theorem to the ab+bc+ca=3 we will achive that abc<=1 (x <= 1). If abc<=1 we can easily prove the inequality 1/a+1/b+1/c >= a+b+c ab+bc+ac >= abc*(a+b+c) a+b+c <= 3/abc a+b+c <= 3/x If we put it to our inequality we will achive 3<= 4*(x^4)-9*(x^3)+15*(x^2)-10*x / 5*(x^2)-4*x-1 4*(x^4)-9*(x^3)+2x+3 >= 0 We can easily prove this inequality, which is right, so the starting inequality was also right.

$ \frac 1{1+a^2(b+c)} + \frac 1{1+b^2(c+a)} + \frac 1 {1+c^2(a+b) } \leq \frac 3 {1+2abc}$ Lemma: $ ab+bc+ca=3$, $ a,b,c \in R^+$ .$ => \frac {6abc}{1+2abc}\leq \frac {((a(b+c)+b(a+c)+c(a+b))^2}{2(a+b+c)+a^2(b+c)^2+b^2(a+c)^2+c^2(a+b)^2}$ . Dem: $ <=> 6abc(2(a+b+c)+a^2(b+c)^2+b^2(a+c)^2+c^2(a+b)^2) \leq 36(1+2abc)$, *$ a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2(abc)(a+b+c)=9-2(abc)(a+b+c)$ $ <=> abc(2(a+b+c)+18-2(abc)(a+b+c) \leq 6(1+2abc)$ $ <=> abc(a+b+c)+3abc-(abc)^2(a+b+c) \leq 3$ $ <=> 0 \leq (3-abc(a+b+c))(abc-1)$..(1) * $ (abc)^{2/3}\leq (ab+bc+ca)/3=1$ $ => abc \leq 1$ => $ abc-1 \leq 0$..(2) * $ xy+yz+zx \leq x^2+y^2+z^2 => 3(xy+yz+zx)\leq (x+y+z)^2$, if $ x=ab, y=bc, z=ca$ =>$ 3(abc)(a+b+c)\leq (ab+bc+ca)^2=9 => (abc)(a+b+c)\leq 3 => abc(a+b+c)-3 \leq 0$..(3) **(3)and (2): $ 0\leq (abc-1)(abc(a+b+c)-3)$=> (1) is true!. $ Q.E.D.$ --- $ S_1=\frac 1{1+a^2(b+c)} + \frac 1{1+b^2(c+a)} + \frac 1 {1+c^2(a+b) }$ $ S_2=\frac 3 {1+2abc}$ $ S_3=\frac {((a(b+c)+b(a+c)+c(a+b))^2}{2(a+b+c)+a^2(b+c)^2+b^2(a+c)^2+c^2(a+b)^2}$ $ 3-S_1=\frac {a^2(b+c)^2}{b+c+a^2(b+c)^2} + \frac {b^2(a+c)^2}{c+a+b^2(c+a)^2} + \frac {c^2(a+b)^2}{a+b+c^2(a+b)^2 }$ $ 3-S_2=\frac {6abc}{1+2abc}$ For Cauchy. $ \frac {(a(b+c)+b(a+c)+c(a+b)^2}{2(a+b+c)+a^2(b+c)^2+b^2(a+c)^2+c^2(a+b)^2} \leq \frac {a^2(b+c)^2}{b+c+a^2(b+c)^2} + \frac {b^2(a+c)^2}{c+a+b^2(c+a)^2} + \frac {c^2(a+b)^2}{a+b+c^2(a+b)^2 }$ For the Lemma. $ 3-S_2 \leq S_3$, but, $ S_3\leq 3-S_1$=> $ 3-S_2\leq 3-S_1$ =>$ S_1\leq S_2$ $ Q.E.D.$

Adding 3 to each side the inequality becomes: $ \sum_{cyc}\frac {a^2(b + c)}{1 + a^2(b + c)} > = \frac {6abc}{1 + 2abc}$, if and only if: $ \sum_{cyc}\frac {(ab + ac)^2}{(b + c) + a^2(b + c)^2} > = \frac {6abc}{1 + 2abc}$, now in order to prove this i am going to use the cauchy inequality giving LHS>=R $ R = \frac {(2ab + 2bc + 2ca)^2}{\sum_{cyc}(b + c) + (ab + ac)^2} = \frac {36}{2(a + b + c) + (3 - bc)^2 + (3 - ca)^2 + (3 - ab)^2}$ $ = \frac {36}{2(a + b + c) + 9 - 6bc + b^2c^2 + 9 - 6ca + c^2a^2 + 9 - 6ab + a^2b^2}$ $ = \frac {36}{2(a + b + c) + 27 - 6(ab + bc + ca) + (a^2b^2 + b^2c^2 + c^2a^2)} = \frac {36}{2(a + b + c) + 9 + (ab + bc + ca)^2 - 2abc(a + b + c)}$ $ = \frac {18}{a + b + c + 9 - abc(a + b + c)}$, and $ R > = \frac {6abc}{1 + 2abc}$, if and only if: $ 3(1 + 2abc) > = abc(a + b + c + 9 - abc(a + b + c))$, letting $ s = a + b + c, p = abc$ $ 3(1 + 2p) > = p(s + 9 - sp)$ if and only if $ 3 - 3p > = ps - sp^2 = ps(1 - p)$ iff $ (1 - p)(3 - ps) > = 0$, this can be easily done by multiplying 2 inequalities: $ \frac {\sqrt{(ab + bc + ca)^3}}{\sqrt{3^3}} > = abc$ by AM-GM inequality, this gives: $ 1 - p > = 0$ $ \frac {(ab + bc + ca)^2}{3} > = (ab)(bc) + (bc)(ca) + (ca)(ab) = abc(a + b + c)$, this is for the well known inequality: $ (p + q + r)^2 > = 3(pq + qr + rp)$, the last inequality is equivalent to $ 3 - ps > = 0$, the problem is done

Let $ P = abc$, and $ S = a + b + c$. Notice $ a^2(b + c) = a(ab + ac) + abc - abc = a(ab + bc + ac) - P = 3a - P$. Similarly, \[ \frac1{1 + a^2(b + c)} + \frac 1{1 + b^2(a + c)} + \frac1{1 + c^2(a + b)} = \frac1{1 + 3a - P} + \frac {1}{1 + 3b - P} + \frac1{1 + 3c - P} \] Lemma A: $ P\leq1$. Consider the given expression $ ab + bc + ac = 3$. It states the arithmetic mean of $ ab$, $ bc$, and $ ac$ is 1. Since $ AM\geq GM$, $ 1\geq (xyz)^{\frac23}$, so $ P\leq1$, equality holding when $ a = b = c = 1$. Lemma B: $ S\geq3$. Using Cauchy on $ < a,b,c >$ and $ < b,c,a >$, $ ab + bc + ac \leq a^2 + b^2 + c^2 \Rightarrow 3(ab + bc + ac) \leq a^2 + b^2 + c^2 + 2(ab + bc + ac) \Rightarrow 9\leq (a + b + c)^2 \Rightarrow 3 \leq S$. Lemma C: $ SP \leq 3$. Let $ p(x) = x^3 - 3x^2 + kx + m$, and let $ a' = ab$, $ b' = bc$, and $ c' = ac$. Let these be roots of the polynomial $ p(x)$. Note $ 3 = a' + b' + c'$, which corresponds to the coefficient of $ x^2$ by Vieta's theorem. Also note $ k = a'b' + b'c' + a'c'$, and that this is equal to $ PS$ (they both reduce to $ a^2bc + ab^2c + abc^2$). $ m = - (a^2b^2c^2)$, so since $ P \leq 1$, $ - 1 \leq m \leq 0$. Now suppose $ k > 3$. Then $ p(x)$ is always increasing, for $ p'(x) > 3x^2 - 6x + 3 = 3(x - 1)^2 \geq0$. But a cubic polynomial (or any function) that is always increasing can only have one real root of multiplicity 1. The other two roots are imaginary, but $ a',b',c' \in \Re$, so we have reached a contradiction. $ k\leq3$, equality holding at $ k = 3$ because $ a^2bc + ab^2c + abc^2 = 3$ when $ a = b = c = 1$, so $ SP \leq 3$. Now we will prove the following inequality: $ 3(P - P^2) + (7P - 1)S \leq 18$ Let's suppose the contrary, that $ 3(P - P^2) + (7P - 1)S > 18$. Then the following lines are implied: \[ 3(P - P^2) + 7PS - S > 18 \] \[ 3(P - P^2) + 7PS - 18 > S \] \[ 3(P - P^2) + PS + 6PS - 18 > S \] \[ 3(P - P^2) + PS > S \quad \quad \quad (PS \leq 3) \] \[ 3P(1 - P) > S(1 - P) \] \[ 3P > S \quad \quad \quad (1 - P \,\texttt{is positive}) \] \[ 3 > S \] We have reached a contradiction, because $ S \geq 3$ because of lemma B. The rest is simply algebra (it may be a little easier to follow this logic backwards; I present it in this order for rigor): \[ 3(P - P^2) + (7P - 1)S \leq 18 \] \[ 9(P - P^2) + 21PS - 3S \leq 54 \] Letting, $ \rho = 1 - P$, and noticing that remarkably $ P - P^2 = \rho - \rho^2$, \[ 9(\rho - \rho^2) + 21S - 21 \rho S - 3S \leq 54 \] \[ 9\rho + 18S \leq 9 \rho ^2 + 21\rho S + 54 \] \[ 9\rho + 18S - 6\rho^2 - 12\rho S - 54 \leq 3\rho^2 + 9\rho S \] \[ 9\rho^2 + 18\rho S - 6\rho^3 - 12\rho^2S - 54\rho + 81\leq 3\rho^3 + 9\rho^2S + 81 \] \[ (3\rho^2 + 6\rho S + 27)(3 - 2\rho) \leq 3(\rho^3 + 3\rho^2S + 27\rho + 27(1 - \rho)) \] \[ (3\rho^2 + 6\rho S + 27)(3 - 2\rho) \leq 3(\rho + 3a)(\rho + 3b)(\rho + 3c) \] \[ \frac {(\rho + 3a)(\rho + 3b) + (\rho + 3a)(\rho + 3c) + (\rho + 3b)(\rho + 3c)}{(\rho + 3a)(\rho + 3b)(\rho + 3c)} \leq \frac3{3 - 2\rho} \] \[ \frac1{\rho + 3a} + \frac1{\rho + 3b} + \frac1{\rho + 3c} \leq \frac 3{3 - 2\rho} \] Substituting back $ 1 - \rho = P$, we get the desired inequality. \[ \frac1{1 + 3a - P} + \frac 1{1 + 3b - P} + \frac1{1 + 3c - P}\leq \frac 3{1 + 2P} \]