Raja Oktovin wrote: Let $ a$, $ b$, and $ c$ be positive real numbers such that $ ab + bc + ca = 3$. Prove the inequality \[ 3 + \sum_{\mathrm{\cyc}} (a - b)^2 \ge \frac {a + b^2c^2}{b + c} + \frac {b + c^2a^2}{c + a} + \frac {c + a^2b^2}{a + b} \ge 3.\]

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I kinda remember that the left side could be changed into the following stronger inequality: \[ 3+\frac12\sum_{cyc}(a-b)^2\geq\sum_{cyc}\frac{a+b^2c^2}{b+c}\] But then I can't prove it. Anybody?

wangsacl wrote: I kinda remember that the left side could be changed into the following stronger inequality: \[ 3 + \frac12\sum_{cyc}(a - b)^2\geq\sum_{cyc}\frac {a + b^2c^2}{b + c}\] But then I can't prove it. Anybody? This ineq is equivalent to :$ {a^2} + {b^2} + {c^2} \ge \sum {\frac {a}{{b + c}} + } \sum {\frac {{{b^2}{c^2}}}{{b + c}}}$ Easy to see that : $ a + b + c \ge 3$ Notice : $ \frac {a}{{b + c}} = \frac {{a(ab + bc + ca)}}{{3(b + c)}} = \frac {{{a^2}}}{3} + \frac {{abc}}{{3(b + c)}} \le \frac {{{a^2}}}{3} + \frac {{a(b + c)}}{{12}}$ And $ \frac {{{b^2}{c^2}}}{{b + c}} \le \frac {{bc(b + c)}}{4}$ And we need to prove that: $ \frac {{2({a^2} + {b^2} + {c^2})}}{3} \ge \frac {1}{2} + \frac {{ab(a + b) + bc(b + c) + ca(c + a)}}{4}$ $ \Leftrightarrow \frac {{2({a^2} + {b^2} + {c^2})}}{3} + \frac {{3abc}}{4} \ge \frac {1}{2} + \frac {{(a + b + c)(ab + bc + ca)}}{4} = \frac {1}{2} + \frac {{3(a + b + c)}}{4}$ By Cchur ineq, we have : $ \frac {1}{4}({a^2} + {b^2} + {c^2} + 3abc) \ge \frac {1}{4}({a^2} + {b^2} + {c^2} + \frac {{9abc}}{{a + b + c}}) \ge \frac {3}{2}$ The proof'll complete if we can show that $ \frac {{5({a^2} + {b^2} + {c^2})}}{{12}} + 1 \ge \frac {{3(a + b + c)}}{4}$ Easy to see that $ LHS \ge \frac {{5{{(a + b + c)}^2}}}{{36}}$ And from the ineq $ \frac {{5{{(a + b + c)}^2}}}{{36}} + 1 \ge \frac {{3(a + b + c)}}{4} \Leftrightarrow (a + b + c - 3)(a + b + c - \frac {{12}}{5}) \ge 0$ ( which is true ) Show that the ineq is true. So we are done !