bboypa wrote: Let $ a,b,c$ positive reals such that $ ab + bc + ca = 3$, show that: $ \displaystyle a^2 + b^2 + c^2 + 3 \ge \frac {a(3 + bc)^2}{(c + b)(b^2 + 3)} + \frac {b(3 + ca)^2}{(a + c)(c^2 + 3)} + \frac {c(3 + ab)^2}{(b + a)(a^2 + 3)}$ (Anass BenTaleb, Ali Ben Bari High School - Taza,Morocco) My proof's below.

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bboypa wrote: Let $ a,b,c$ positive reals such that $ ab + bc + ca = 3$, show that: $ \displaystyle a^2 + b^2 + c^2 + 3 \ge \frac {a(3 + bc)^2}{(c + b)(b^2 + 3)} + \frac {b(3 + ca)^2}{(a + c)(c^2 + 3)} + \frac {c(3 + ab)^2}{(b + a)(a^2 + 3)}$ (Anass BenTaleb, Ali Ben Bari High School - Taza,Morocco) Second proof: We have \[ \frac{a(bc+3)^2}{(b+c)(b^2+3)}=\frac{a(2bc+a(b+c))^2}{(a+b)(b+c)^2} =\frac{a[4bc(ab+bc+ca)+a^2(b+c)^2]}{(a+b)(b+c)^2} \le \frac{a[(b+c)^2(ab+bc+ca)+a^2(b+c)^2]}{(a+b)(b+c)^2} =a(a+c)\] Using this for the similar terms and adding them up, we can get the result.

@Honey_S, you uses only am-gm, perfect @can_hang2007, very nice too

Let $ x=\frac{1}{\sqrt{3}}a,y=\frac{1}{\sqrt{3}}b,z=\frac{1}{\sqrt{3}}c$ then $ xy+yz+zx=1$ and $ x=\frac{1-zy}{z+y}$ (1) and we must prove $ \sum x^2+1\ge \sum \frac{(1+yz)^2(1-yz)}{(1-yz)^2+(y+z)^2}$ (we only use (1) ) we will prove $ \frac{(1+yz)^2}{(1-yz)^2+(y+z)^2}\le 1$ $ 4zy\le (z+y)^2$ and its obviously true by AM-GM so $ \sum \frac{(1+yz)^2(1-yz)}{(1-yz)^2+(y+z)^2}\le \sum (1-yz)=1+xy+yz+zx\le \sum x^2+1$ (last true by CS) so we are done. Equality when x=y=z or a=b=c=1

Sorry for not participating to the contest. I just opened my mailbox after 10 days and I just realized that I lost the final round My solution to the 1st problem: By Cauchy-Swartz we have: $ LHS = \sum_{cyclic}{\frac {a(\sqrt {3}\sqrt {3} + bc)^2}{(c + b)(b^2 + 3)}} \leq \sum_{cyclic}{\frac {a(b^2 + 3)(c^2 + 3)}{(c + b)(b^2 + 3)}} = \sum_{cyclic}{\frac {a(c^2 + 3)}{(c + b)}} = \sum_{cyclic}{\frac {a(c^2 + ab + bc + ca)}{(c + b)}} = \sum_{cyclic}{\frac {a(c + b)(c + a)}{(c + b)}} = \sum_{cyclic}{a(c + a)} = a^2 + b^2 + c^2 + ab + bc + ca = a^2 + b^2 + c^2 + 3$