The equality holds when $a=b=c=\frac{1}{\sqrt{3}}$.

In this case we have: $\frac{1}{a^2-bc+1}\leq1\Leftrightarrow a^2-bc+1\geq1$ $\frac{1}{b^2-ac+1}\leq1\Leftrightarrow b^2-ac+1\geq1$ $\frac{1}{c^2-ab+1}\leq1\Leftrightarrow c^2-ab+1\geq1$ Summing the 3 relations $\frac{1}{a^2-bc+1}+\frac{1}{b^2-ac+1}+\frac{1}{c^2-ab+1}\leq3\Leftrightarrow a^2+b^2+c^2\geq\frac{1}{3}$. If $a=b=c$ it's ok , else... i don't know

Slizzel wrote: In this case we have: $\frac{1}{a^2-bc+1}\leq1\Leftrightarrow a^2-bc+1\geq1$ I can't understand.

The fraction $\frac{1}{a^2-bc+1}$ to less (or equal) then 1 it's denominator should be $\geq1$

I think your solution is wrong

Slizzel wrote: In this case we have: $\frac{1}{a^2-bc+1}\leq1\Leftrightarrow a^2-bc+1\geq1$ $\frac{1}{b^2-ac+1}\leq1\Leftrightarrow b^2-ac+1\geq1$ $\frac{1}{c^2-ab+1}\leq1\Leftrightarrow c^2-ab+1\geq1$ Summing the 3 relations $\frac{1}{a^2-bc+1}+\frac{1}{b^2-ac+1}+\frac{1}{c^2-ab+1}\leq3\Leftrightarrow a^2+b^2+c^2\geq\frac{1}{3}$. If $a=b=c$ it's ok , else... i don't know Sorry but this is totally wrong : It's true that if we have $\frac{1}{a^{2}-bc+1}\leq 1$ and $\frac{1}{b^{2}-ca+1}\leq 1$ and $\frac{1}{c^{2}-ab+1}\leq 1$ this means that $\frac{1}{a^{2}-bc+1}+\frac{1}{b^{2}-ca+1}+\frac{1}{c^{2}-ab+1}\leq 3$ But if we had $\frac{1}{a^{2}-bc+1}+\frac{1}{b^{2}-ca+1}+\frac{1}{c^{2}-ab+1}\leq 3$ It doesn't necesserly mean that $\frac{1}{a^{2}-bc+1}\leq 1$ and $\frac{1}{b^{2}-ca+1}\leq 1$ and $\frac{1}{c^{2}-ab+1}\leq 1$

our ineq is equivalent to: $\sum\frac{a(a+b+c)}{3a(a+b+c)+2}\geq\frac{1}{3}$ what can we do ????????/ let $x=a(a+b+c)$ why not??????????

Homogenizing, the ineq is equivalent to \[ \sum_\mathrm{cyc} \frac{1}{a(a+b+c)+2(ab+bc+ca)} \leq \frac{1}{ab+bc+ca} \] Multiply both sides by $2(ab+bc+ca)$, \[ \sum_\mathrm{cyc} \frac{2(ab+bc+ca)}{a(a+b+c)+2(ab+bc+ca)} \leq 2 \] Subtracting from 3, the above ineq is equivalent to \[ \sum_\mathrm{cyc} \frac{a(a+b+c)}{a(a+b+c)+2(ab+bc+ca)} \geq 1 \] Now Cauchy \[ LHS = (a+b+c)\sum_\mathrm{cyc} \frac{a^2}{a^2(a+b+c)+2a(ab+bc+ca)} \geq \frac{(a+b+c)^3}{\sum_\mathrm{cyc} [a^2(a+b+c)+2a(ab+bc+ca)] } = 1 \]

Very nice!

hei, very nice!! and this is my proof: our ineq is equyvalent to: $\sum\frac{a}{3a(a+b+c)+2}\geq\frac{\frac{1}{3}}{a+b+c}$ by cauchy: $lhs\geq\frac{(a+b+c)^2}{3(a+b+c)(a^2+b^2+c^2)+2(a+b+c)}=\frac{\frac{1}{3}}{a+b+c}$ which is true: $\frac{a+b+c}{3(a^2+b^2+c^2)+2}=\frac{\frac{1}{3}}{a+b+c}$ multifly we get $ab+bc+ca=1/3$ it's cost me alot of time

From the moment I saw this inequality, the next thing that came into my mind was this inequality proposed by Titu Andreescu in G.M in 2001. Let $a, b, c>0$ such that $ab+bc+ca=\frac{1}{3}.$ Prove that: \[\frac{a}{a^{2}-bc+1}+\frac{b}{b^{2}-ca+1}+\frac{c}{c^{2}-ab+1}\geq\frac{1}{a+b+c}. \] Solution. Let us apply the well-known inequality Cauchy-Schwarz. We have: \[(a(a^{2}-bc+1)+b(b^{2}-ca+1)+c(c^{2}-ab+1))(\frac{a}{a^{2}-bc+1}+\frac{b}{b^{2}-ca+1}+\frac{c}{c^{2}-ab+1})\geq (a+b+c)^{2}, \] so we get that \[\frac{a}{a^{2}-bc+1}+\frac{b}{b^{2}-ca+1}+\frac{c}{c^{2}-ab+1}\geq\frac{(a+b+c)^{2}}{a^{3}+b^{3}+c^{3}-3abc+a+b+c}. \] Now we only need to prove that $\frac{(a+b+c)^{2}}{a^{3}+b^{3}+c^{3}-3abc+a+b+c}\geq\frac{1}{a+b+c}.$ But this inequality is in fact equality because using the well-known identity $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$, we deduce that $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}-ab-bc-ca,$ so we are done. Now let's deduce this chinese problem from this one: We have: \[\sum\frac{a}{a^{2}-bc+1}\geq\frac{1}{a+b+c}\] But this is wquivalent with: \[\sum\frac{a^{2}+ab+ac}{a^{2}-bc+1}\geq 1, \] or \[\sum\frac{(a^{2}-bc+1)+(ab+bc+ca-)}{a^{2}-bc+1}\geq 1 \] or \[3+\sum\frac{ab+bc+ca-1}{a^{2}-bc+1}\geq 1 \] or \[\sum\frac{1}{a^{2}-bc+1}\leq 3. \]

when i see this kind of inequalities i know that almost everytime there a "nice" solution but i do not have the time to find it so i find only stupid solutions. i will only say the main steps since i do not have the time to post a full solution right now.denote $x=ab,y=bc,z=ca \Rightarrow a^2=xz/y.b^2=xy/z,c^2=yz/x$.then the conditions rewrites as $x+y+z=1/3$ and our ineq rewrites as $3\ge y/(xz-y^2+y) +.......)$.now we put $x=a/3(a+b+c),y=.....$(note that a,b,c are different from the initial ones-you may also put $a_1,a_2,a_3$ instead of $a,b,c...$) finally our ineq is equivalent with $1\ge b(a+b+c)/(ac-b^2+3b(a+b+c))+.........$ and that is also equivalent with $(ac-b^2)/(ac+3ab+3bc+b^2)+.....\ge0$ and that you easily kill be clearing everything out and then some muerhead

The inequality is equivalent to 1/(a*b+b*c+c*a)-1/(a^2-b*c+3*(a*b+b*c+c*a))-1/(b^2-c*a+3*(a*b+b*c+c*a))-1/(c^2-a*b+3*(a*b+b*c+c*a))>=0 <==> -3*a^2*b^2*c^2-b^2*c^3*a+3*b*c^4*a-b^3*c^2*a+3*b^4*c*a-a^2*b*c^3-a^3*c^2*b+3*a^4*c*b-a^2*b^3*c -a^3*b^2*c-2*a^3*b^3-2*a^3*c^3+a^4*c^2+a^4*b^2-2*b^3*c^3+b^4*c^2+b^2*c^4+a^2*b^4+c^4*a^2 =F[6][0,3]+5*F[6][1,1]+2*F[6][1,2]>=0, in which F[6][0,3]=sum(a^2*(b-c)^2*(a-b)*(a-c))>=0; F[6][1,1]=sum(a^2*b*c*(a-b)*(a-c))>=0; F[6][1,2]=sum(a*b*c*(b+c)*(a-b)*(a-c))>=0.

$\sum_{cyc}\frac{1}{a^{2}-bc+1}\leq3\Leftrightarrow\sum_{cyc}\frac{ab+ac+bc}{a^{2}-bc+1}\leq1\Leftrightarrow$ $\Leftrightarrow\sum_{cyc}\left(1-\frac{2(ab+ac+bc)}{a^{2}-bc+1}\right)\geq1\Leftrightarrow\sum_{cyc}\frac{a^{2}+ab+ac}{a^{2}-bc+1}\geq1.$ But $\sum_{cyc}\frac{a^{2}+ab+ac}{a^{2}-bc+1}=$ $=\sum_{cyc}\frac{a^{2}(a+b+c)}{a^{3}-abc+a}\geq\frac{(a+b+c)^{3}}{a^{3}+b^{3}+c^{3}-3abc+a+b+c}=1.$

As a note, I'd like to point out that this inequality can be "dumbassed". The desired inequality is equivalent to $ \sum_{cyc}\frac{ab+ac+bc}{a^2+3ab+3ac+2bc} \leq 1$, which is equivalent to $ (ab+ac+bc)\sum_{cyc}(a^2+3ab+3ac+2bc)(b^2+3bc+3ab+2ac) \leq (a^2+3ab+3ac+2bc)(b^2+3bc+3ab+2ac)(c^2+3ac+3bc+2ab)$, or, expanding $ 5\sum_{sym}a^4b^2 + 11\sum_{sym}a^3b^3 + 5\sum_{sym}a^4bc + 76\sum_{sym}a^3b^2c + 147a^2b^2c^2 \leq 6\sum_{sym}a^4b^2 + 10\sum_{sym}a^3b^3 + \frac{13}{2}\sum_{sym}a^4bc + 75\sum_{sym}a^3b^2c + 144a^2b^2c^2$. Cancelling terms, we get the equivalent $ \sum_{sym}a^3b^3 + \sum_{sym}a^3b^2c + 3a^2b^2c^2 \leq \sum_{sym}a^4b^2 + \frac{3}{2}\sum_{sym}a^4bc$, which is evident by weighted AM-GM.

I think the inequality can also be done with the SOS method. Let $ X = a + b + c$ and $ Y = ab + ac + bc$. The inequality can be rewritten as (through some manipulation) $ \sum (a - b)^2(2Y^2 + \frac {X^2}{2}(ac + bc - ab))$ or $ \sum (a - b)^2S_c$ for the $ S_a, S_b, S_c$ chosen as above. WLOG, $ a\geq b \geq c$, so it is clear that $ S_a$, $ S_b, S_b + S_c \geq 0$, and that $ (a - c) \geq (a - b)^2$ so we are done.

China TST 2005

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Macedonia National Olympiad 2009 Macedonia National Olympiad 2009 - Problem 4 The following inequality is also true. Let $a$, $b$, $c$ be nonnegative reals such that $\sqrt{ab}+\sqrt{bc}+\sqrt{ca} = \frac{1}{3}$. Prove that \[\frac{1}{a^{2}-bc+1}+\frac{1}{b^{2}-ca+1}+\frac{1}{c^{2}-ab+1}\leq 3 .\]

billzhao wrote: Let $a$, $b$, $c$ be nonnegative reals such that $ab+bc+ca = \frac{1}{3}$. Prove that \[\frac{1}{a^{2}-bc+1}+\frac{1}{b^{2}-ca+1}+\frac{1}{c^{2}-ab+1}\leq 3 \] So here's my long solution and probably brainless the inequality can be rewritten as $$\sum_{a,b,c}\frac{a^2-bc}{a^2-bc+1} \ge 0$$$$\iff \sum_{a,b,c} (a^2-bc)(b^2ca+1)(c^2-ab+1) \ge 0$$$$\iff \sum_{a,b,c} (a^2-bc)(b^2c^2-b^3a+b^2-c^3a+a^2bc-ca+c^2 -ab) \ge 0$$$$\iff \sum_{a,b,c} (-a^3b^3-a^3c^3-b^3c^3 + a^2b^2+a^2c^2 +abc(a^2+b^2+c^3)-a^3b-a^3c-b^3c-bc^3 +a^2 -bc +abc(b+c)) \ge 0$$ Setting $a+b+c = u$, $v=ab+bc+ca=\frac{1}{3}$, $w=abc$ It can be easily seen $a^3b^3+b^3c^3+c^3a^3 = v(v^2-3uw)+3w^2$ and $\sum_{a,b,c}(a^3b+a^3c+b^3c+bc^3) = 2(u^2v-2v^2-uw)$ Replacing this and other values, the inequality reduces to $$\iff -3(v(v^2-3uw)+3w^2)+2(v^2-2uw)+3w(u^3-3uv+3w) - 2(u^2v-2v^2-uw) + u^2 - 2v -v +2uw \ge 0$$$$ \iff -3v^3+6v^2+3u^3w-2u^2v+u^2-3v \ge 0$$ Replacing $v= \frac{1}{3}$ $$\iff 3u^2+27u^3w \ge 4$$ Now there are 2 cases. Case 1 : $a,b,c$ form sides of a triangle Then using Schur's inequality, $a^3+b^3+c^3 + 3abc \ge ab(a+b) + bc(b+c) + ca(c+a)$ $\implies u^3+9w \ge uv$ $\iff 27w \ge 4u-3u^3$ $3u^3 + 27u^3w - 4 \ge 3u^2 + u^3(4u-3u^3) - 4 = (3u^2 - 4)(1-u^4)$ Since $a,b,c$ form sides of a triangle, $3u^2 - 4 \le 0$ and obviously $1-u^4 \le 0$ $$\implies 3u^3 + 27u^3w \ge 4$$QED Case 2 : $a,b,c$ don't form the sides of a triangle. Normalizing, we are left to prove $$\iff u^2v^2 + u^3w \ge 4v^3$$Since $a,b,c$ are not the sides of a triangle, Without loss of generality, $$a \ge b+c \ge \frac{u}{2}$$$$u^3a \ge \frac{u^4}{2} \ge 4.5 v^2 \ge 4v^2$$$$\implies u^3w \ge v^24bc \ge v^2(4bc - (b+c-a)^2)$$$$\implies u^2v^2 + u^3w \ge 4v^3$$ QED

billzhao wrote: Let $a$, $b$, $c$ be nonnegative reals such that $ab+bc+ca = \frac{1}{3}$. Prove that \[\frac{1}{a^{2}-bc+1}+\frac{1}{b^{2}-ca+1}+\frac{1}{c^{2}-ab+1}\leq 3 \] We have \[3-\sum{\frac{1}{a^2-bc+1}}=\frac{\sum{ab(2+3c^2+3bc+3ca)(a-b)^2}}{2(a^2-bc+1)(b^2-ca+1)(c^2-ab+1)}\ge{0}\]

Kunihiko_Chikaya wrote: The equality holds when $a=b=c=\frac{1}{\sqrt{3}}$. Then, ab + bc + ca = 1, not $\frac{1}{3}.$

arqady wrote: $\sum_{cyc}\frac{1}{a^{2}-bc+1}\leq3\Leftrightarrow\sum_{cyc}\frac{ab+ac+bc}{a^{2}-bc+1}\leq1\Leftrightarrow$ $\Leftrightarrow\sum_{cyc}\left(1-\frac{2(ab+ac+bc)}{a^{2}-bc+1}\right)\geq1\Leftrightarrow\sum_{cyc}\frac{a^{2}+ab+ac}{a^{2}-bc+1}\geq1.$ But $\sum_{cyc}\frac{a^{2}+ab+ac}{a^{2}-bc+1}=$ $=\sum_{cyc}\frac{a^{2}(a+b+c)}{a^{3}-abc+a}\geq\frac{(a+b+c)^{3}}{a^{3}+b^{3}+c^{3}-3abc+a+b+c}=1.$ Nice

First, we homogenize to get \[\frac{1}{a^2 - bc + 1} = \frac{3(ab+bc+ca)}{a^2 + 3ab +3ac + 2ab} = \frac{3}{2} - \frac{\frac{3}{2}(a)(a+b+c)}{a(a+b+c)+2(ab+bc+ca)}.\]Substituting this expression into the inequality means that it suffices to prove \[\iff \sum_{cyc}\frac{a}{a(a+b+c)+2(ab+bc+ca)} \geq \frac{1}{a+b+c}.\]By Cauchy-Schwarz, we have \begin{align*} \sum_{cyc}\frac{a}{a(a+b+c)+2(ab+bc+ca)} &\overset{\text{Cauchy}}{\geq} \frac{(a+b+c)^2}{\sum_{cyc} a^2(a+b+c) + 2a(ab+bc+ca)} \\ &\geq \frac{(a+b+c)^2}{(a^2+b^2+c^2)(a+b+c) + 2(a+b+c)(ab+bc+ca)} \\ &\geq \frac{(a+b+c)^2}{(a+b+c)(a^2+b^2+c^2 + 2ab+2bc+2ca)} \\ &\geq \frac{1}{a+b+c}. \end{align*}

I will present a bashy solution. Let $a=\frac{x}{3},b=\frac{y}{3},c=\frac{z}{3}$. \[\sum{\frac{1}{x^2+xy+xz+6}}=\sum{\frac{1}{x^2-yz+9}}\overset{?}{\leq} \frac{1}{3}\]Let $x+y+z=p$. Now we will expand this inequality. \[\sum{\frac{1}{px+6}}\overset{?}{\leq} \frac{1}{3}\iff 3(3p^2+12p^2+108)\overset{?}{\leq} p^3xyz+18p^2+36p^2+216\iff (x+y+z)^2(9+xyz(x+y+z))\overset{?}{\geq }108\]\[6\sum{x}+9\sum{x^2}+xyz\sum{x}\sum{x^2}\overset{?}{\geq} 54\]Let $x+y+z=3u,xy+yz+zx=3v^2,xyz=w^3$. \[12\sum{x}+27\sum{x^2}\overset{?}{\geq} 108+\sum{x^2}\sum{x^2y^2}\]By $\sum{x^2y^2}+2xyz\sum{x}=9$. Let's homogenise the inequality in order to use $uvw$. \[3(\sum{xy})^2\sum{x^2}\overset{?}{\geq} (\sum{xy})^3.\frac{8}{3}+\sum{x^2}\sum{x^2y^2}\]\[27v^4(9u^2-6v^2)\overset{?}{\geq} 10v^6-9u^2v^4\iff 72v^6+(9u^2-6v^2)(9v^4-6uw^3)\iff 162u^2v^454u^3w^3\overset{?}{\geq} 180v^6+36uv^2w^3\]\[\iff 9u^2v^4+3u^3w^3\overset{?}{\geq} 10v^6+2uv^2w^3\]Note that by Schur with $3$rd degree, $w^3\geq 4uv^2-3u^3$. Since $u\geq v$, $3u^3-2uv^2>0$. Hence \[w^3(3u^3-2uv^2)\overset{?}{\geq} 10v^6-9u^2v^4\iff w^3\overset{?}{\geq} \frac{v^4(10v^2-9u^2)}{u(3u^2-2v^2)}\]If $9u^2\geq 10v^2$, then $w^3\geq 0\geq RHS$. Assume that $9u^2<10v^2$. \[w^3\geq 4uv^2-3u^3\overset{?}{\geq} \frac{v^4(10v^2-9u^2)}{u(3u^2-2v^2)}\]We have $u(4v^2-3u^2)(3u^2-2v^2)\overset{?}{\geq} v^4(10v^2-9u^2)$ and this is true since $9u^2<10v^2$ implies $3u^2<4v^2$, we see that $u^2(4v^2-3u^2)(3u^2-2v^2)\geq u^4(4v^2-3u^2)\geq v^4(4v^2-3u^2)\geq v^4(10v^2-9u^2)$ as desired.$\blacksquare$

First homogenize to get that the ineq is $\sum_{cyc} \frac{ab+bc+ca}{a(a+b+c)+2(ab+bc+ac)} \le 1$, now multiply everything by $2$ and then add $1$ on each side and subtract the fractions to get that now we want to prove that $S=\sum_{cyc} \frac{a(a+b+c)}{a(a+b+c)+2(ab+bc+ac)} \ge 1$. \[S=(a+b+c) \sum_{cyc} \frac{a^2}{a^2(a+b+c)+2a(ab+bc+ca)} \overset{T2}{\ge} \frac{(a+b+c)^3}{a^3+b^3+c^3+6abc+3 \sum_{sym} a^2b}=1 \]Therefore we are done .