MJ GEO wrote: $ a^2 + b^2 + c^2 = 1$.prove that: $ \frac {a}{b^2 + 1} + \frac {b}{c^2 + 1} + \frac {c}{a^2 + 1} \geq \frac {3}{4}(a\sqrt {a} + b\sqrt {b} + c\sqrt {c})$ This is problem from - Hojoo Lee "Topics in ineq" or Greece olympiad

Its from Mediterranean 2002.

MJ GEO wrote: Its from Mediterranean 2002. But it's true.... But I don't solution

from Cauchy ineq we get that $ (x_1^2 + x_2^2 + x_3^2)(y_1^2 + y_2^2 + y_3^2)\geq(x_1y_1 + x_2y_2 + x_3y_3)^2$ $ x_1 = \sqrt {a^2(b^2 + 1)}$ $ x_2 = \sqrt {b^2(c^2 + 1)}$ $ x_3 = \sqrt {c^2(a^2 + 1)}$ $ y_1 = \sqrt\frac {a}{b^2 + 1}$ , $ y_2 = \sqrt\frac {b}{c^2 + 1}$ , $ y_3 = \sqrt\frac {c}{a^2 + 1}$ so we get $ [a^2(b^2 + 1) + b^2(c^2 + 1) + c^2(a^2 + 1)][\frac {a}{b^2 + 1} + \frac {b}{c^2 + 1} + \frac {c}{a^2 + 1}]\geq (a\sqrt {a} + b\sqrt {b} + c\sqrt {c})^2$ (1) so we only need to prove that $ \frac {3}{4}\geq a^2(b^2 + 1) + b^2(c^2 + 1) + c^2(a^2 + 1)$ (2) (1),(2) we get that $ \frac {a}{b^2 + 1} + \frac {b}{c^2 + 1} + \frac {c}{a^2 + 1} \geq \frac {(a\sqrt {a} + b\sqrt {b} + c\sqrt {c})^2}{a^2(b^2 + 1) + b^2(c^2 + 1) + c^2(a^2 + 1)} \geq \frac {3}{4}(a\sqrt {a} + b\sqrt {b} + c\sqrt {c})^2$ (2) equals $ \frac {3}{4}\geq a^2b^2 + a^2 + b^2c^2 + b^2 + c^2a^2 + c^2$ $ \frac {1}{3} \geq a^2b^2 + b^2c^2 + c^2a^2$ and meacause $ a^2 + b^2 + c^2 = 1$ "easy " we get that $ \frac {1}{2}((a^2 - b^2)^2 + (b^2 - c^2)^2 + (c^2 - a^2)^2) \geq 0$ so $ a^2 = b^2 = c^2 = \frac {1}{3}$ so $ a = b = c = \frac {\sqrt {3}}{3}$ and beacause then the vectors $ (x_1,x_2,x_3)$ and $ (y_1,y_2,y_3)$ is collinear ( i dont know if that the right word sorry for my english at all ) so (2) = $ \frac {4}{3}$

Sorry to revive an oldish thread, but Mediterranean 2002 has only just been added here. Am I right in thinking the question should be to prove that: $\frac {a}{b^2 + 1} + \frac {b}{c^2 + 1} + \frac {c}{a^2 + 1} \geq \frac {3}{4}(a\sqrt {a} + b\sqrt {b} + c\sqrt {c})^2$ .....with the bracket on the right SQUARED ?? Also I don't think KapioPulsar's last stage is completely correct: KapioPulsar wrote: $\frac {1}{3} \geq a^2b^2 + b^2c^2 + c^2a^2 $ and meacause $a^2 + b^2 + c^2 = 1$ "easy " we get that $\frac {1}{2}((a^2 - b^2)^2 + (b^2 - c^2)^2 + (c^2 - a^2)^2) \geq 0$ This shows that $a^2b^2 + b^2c^2 + c^2a^2 \leq a^4+b^4+c^4 \geq 1/3$ which is not the same as showing what you want, i.e. that $a^2b^2 + b^2c^2 + c^2a^2 \leq 1/3$ It seems clearer to say $a^2b^2 + b^2c^2 + c^2a^2 \leq 1/3(a^2+b^2+c^2)^2 = 1/3$

Merlinaeus wrote: Sorry to revive an oldish thread, but Mediterranean 2002 has only just been added here. Am I right in thinking the question should be to prove that: $\frac {a}{b^2 + 1} + \frac {b}{c^2 + 1} + \frac {c}{a^2 + 1} \geq \frac {3}{4}(a\sqrt {a} + b\sqrt {b} + c\sqrt {c})^2$ .....with the bracket on the right SQUARED ?? Also I don't think KapioPulsar's last stage is correct: KapioPulsar wrote: $\frac {1}{3} \geq a^2b^2 + b^2c^2 + c^2a^2 $ and meacause $a^2 + b^2 + c^2 = 1$ "easy " we get that $\frac {1}{2}((a^2 - b^2)^2 + (b^2 - c^2)^2 + (c^2 - a^2)^2) \geq 0$ This shows that $a^2b^2 + b^2c^2 + c^2a^2 \leq a^4+b^4+c^4 \geq 1/3$ which is not the same as showing what you want, i.e. that $a^2b^2 + b^2c^2 + c^2a^2 \leq 1/3$ Thanks for correction. Edited.