the inequality is equivalent to : $ \sum\frac{(ab)^2}{c^3(a^2-ab+b^2)}\geq \frac{3(a+b+c)}{ab+ac+bc}$ $ \iff$ $ \sum\frac{(ab)^5}{a^2-ab+b^2}\geq \frac{3(abc)^3(a+b+c)}{ab+ac+bc}$ but : $ 3(abc)^3(a+b+c)=3abc(\sum a)(abc)^2\leq (\sum ab)^2(abc)^2$ (AM-GM) hence it suffices to prove that : $ \sum\frac{(ab)^5}{a^2-ab+b^2}\geq (abc)^2(\sum ab)$ $ \iff$ $ \sum\frac{(ab)^3}{c^2(a^2-ab+b^2)}\geq \sum ab$ $ \iff$ $ \sum\frac{(ab)^3}{c^2(a^2-ab+b^2)}+\sum c(a+b)\geq 3\sum ab$ $ \iff$ $ \sum\frac{(ab)^3+(bc)^3+(ca)^3}{c^2(a^2-ab+b^2)}\geq 3\sum ab$ but $ \sum\frac{(ab)^3+(bc)^3+(ca)^3}{c^2(a^2-ab+b^2)}\geq 9\frac{(ab)^3+(ac)^3+(bc)^3}{2\sum(ab)^2-abc(\sum a)}$ it suffices to prove that : $ 9\frac{(ab)^3+(ac)^3+(bc)^3}{2\sum(ab)^2-abc(\sum a)}\geq 3\sum ab$ denote $ x=ab , y=ac$ and $ z=bc$ $ 3\frac{x^3+y^3+z^3}{2(x^2+y^2+z^2)-xy+yz+zx}\geq x+y+z$ $ \iff$ $ \sum x^3+3xyz\geq \sum xy(x+y)$ wich is schur inequality ,and we have done

mashel wrote: Let a.b.c be positive reals such that their sum is 1. Prove that $ \frac {a^{2}b^{2}}{c^{3}(a^{2} - ab + b^{2})} + \frac {b^{2}c^{2}}{a^{3}(b^{2} - bc + c^{2})} + \frac {a^{2}c^{2}}{b^{3}(a^{2} - ac + c^{2})}\geq \frac {3}{ab + bc + ac}$ The inequality is equivalent to \[ \sum \frac{a^2b^2}{c^3(a^2-ab+b^2)} \ge \frac{3(a+b+c)}{ab+bc+ca}\] Put $ x=\frac{1}{a},y=\frac{1}{b},z=\frac{1}{c}$, then the above inequality becomes \[ \sum \frac{z^3}{x^2-xy+y^2} \ge \frac{3(xy+yz+zx)}{x+y+z}\] This is a very known inequality

Using CS Engel, we get that $ (\sum ab)^3 \ge 3 \sum c^3(a^2-ab+b^2)$ which can be proved by simple manipulation.

Agr_94_Math wrote: Using CS Engel, we get that $ (\sum ab)^3 \ge 3 \sum c^3(a^2 - ab + b^2)$ which can be proved by simple manipulation. I'd like to see how this can be be proved by "simple manipulation" : @can_hang2007, Could you please give me a link to your known inequality.

can_hang2007 wrote: mashel wrote: Let a.b.c be positive reals such that their sum is 1. Prove that $ \frac {a^{2}b^{2}}{c^{3}(a^{2} - ab + b^{2})} + \frac {b^{2}c^{2}}{a^{3}(b^{2} - bc + c^{2})} + \frac {a^{2}c^{2}}{b^{3}(a^{2} - ac + c^{2})}\geq \frac {3}{ab + bc + ac}$ The inequality is equivalent to \[ \sum \frac {a^2b^2}{c^3(a^2 - ab + b^2)} \ge \frac {3(a + b + c)}{ab + bc + ca} \] Put $ x = \frac {1}{a},y = \frac {1}{b},z = \frac {1}{c}$, then the above inequality becomes \[ \sum \frac {z^3}{x^2 - xy + y^2} \ge \frac {3(xy + yz + zx)}{x + y + z} \] This is a very known inequality I will prove the inequality than stronger: $ \sum \frac {z^3}{x^2-xy+y^2} \ge\ x+y+z$ BY CAUCHY-SCHWARZ inequality ,we have: $ LHS \ge\ \frac{(x^2+y^2+z^2)^2}{\sum{z(x^2-xy+y^2)}}$ So, we must prove that: $ (x^2+y^2+z^2)^2\ge\ (x+y+z)({\sum{z(x^2-xy+y^2)})}$ $ <=>x^4+y^4+z^4+xyz(x+y+z) \ge\ x^3(y+z)+y^3(x+z)+z^3(x+y)$ Which is obvious true by Schur inequality

Quote: Agr_94_Math wrote: Using CS Engel, we get that .. I'm sorry,what is CS Engel :

CS engel is Cauchy-Schwartz-Bunjakovsky inequality which clears the denominators,sorry for not being specific,but i don't know how to latex that form.

mashel wrote: Let a.b.c be positive reals such that their sum is 1. Prove that $ \frac {a^{2}b^{2}}{c^{3}(a^{2} - ab + b^{2})} + \frac {b^{2}c^{2}}{a^{3}(b^{2} - bc + c^{2})} + \frac {a^{2}c^{2}}{b^{3}(a^{2} - ac + c^{2})}\geq \frac {3}{ab + bc + ac}$ I have two inequality Pro. let $ a,b,c > 0, a+b+c=1$ and real number $ k$. Prove that: 1/ $ \frac {{{a^{k - 1}}{b^{k - 1}}(a + b)}}{{{c^k}\left( {{a^k} + {b^k}} \right)}} + \frac {{{b^{k - 1}}{c^{k - 1}}(b + c)}}{{{a^k}\left( {{b^k} + {c^k}} \right)}} + \frac {{{c^{k - 1}}{a^{k - 1}}(c + a)}}{{{b^k}\left( {{c^k} + {a^k}} \right)}} \ge \frac {3}{{ab + bc + ca}}$ with $ k\geq 1$ or $ k < 0$ 2/ $ \frac {{{a^{k - 1}}{b^{k - 1}}(a + b)}}{{{c^k}\left( {{a^k} + {b^k}} \right)}} + \frac {{{b^{k - 1}}{c^{k - 1}}(b + c)}}{{{a^k}\left( {{b^k} + {c^k}} \right)}} + \frac {{{c^{k - 1}}{a^{k - 1}}(c + a)}}{{{b^k}\left( {{c^k} + {a^k}} \right)}} \le \frac {1}{{3abc}}$ with $ 1 \ge k \ge 0$

First we multiply $ a + b + c$ Right side and we have $ \frac {a^{2}b^{2}}{c^{3}(a^{2} - ab + b^{2})} + \frac {a^{2}c^{2}}{b^{3}(a^{2} - ac + c^{2})} + \frac {b^{2}c^{2}}{a^{3}(b^{2} - bc + c^{2})}\geq \frac {3(a + b + c)}{(ab + ac + bc)}$ This inequality is true for all a,b,c if and only if main inequality is true for $ a + b + c = 1$ This inequality homogeneous and all degree are equal we can $ abc = 1$ $ abc = 1$ $ \implies$ $ \frac {a^{5}b^{5}}{a^{2} - ab + b^{2}} + \frac {b^{5}c^{5}}{b^{2} - bc + c^{2}} + \frac {a^{5}c^{5}}{a^{2} - ac + c^{2}}$ $ \implies$ $ a^{5} + b^{5}\geq ab(a^{3} + b^{3})$ similarly we have $ ab(a + b) + bc(b + c) + ac(a + c)\geq\frac {3(a + b + c)}{ab + ac + bc}$ from Muirhead inequality $ (2,1,0)(1,1,0)\geq 6(1,0,0)$ $ \Box$

mestav wrote: First we multiply $ a + b + c$ Right side and we have $ \frac {a^{2}b^{2}}{c^{3}(a^{2} - ab + b^{2})} + \frac {a^{2}c^{2}}{b^{3}(a^{2} - ac + c^{2})} + \frac {b^{2}c^{2}}{a^{3}(b^{2} - bc + c^{2})}\geq \frac {3(a + b + c)}{(ab + ac + bc)}$ This inequality is true for all a,b,c if and only if main inequality is true for $ a + b + c = 1$ This inequality homogeneous and all degree are equal we can $ abc = 1$ $ abc = 1$ $ \implies$ $ \frac {a^{5}b^{5}}{a^{2} - ab + b^{2}} + \frac {b^{5}c^{5}}{b^{2} - bc + c^{2}} + \frac {a^{5}c^{5}}{a^{2} - ac + c^{2}}$ $ \implies$ $ a^{5} + b^{5}\geq ab(a^{3} + b^{3})$ similarly we have $ ab(a + b) + bc(b + c) + ac(a + c)\geq\frac {3(a + b + c)}{ab + ac + bc}$ from Muirhead inequality $ (2,1,0)(1,1,0)\geq 6(1,0,0)$ $ \Box$ Mestav I don't understand ,because $ a^{5} + b^{5}\geq ab(a^{3} + b^{3})$ it is true But $ a^{5}b^{5}\geq ab(a^{3}+b^{3})$ is it true

Sorry i made huge mistake but we could prove following ineq.for $ abc=1$ easily with muirhead inequality $ \frac{a^{5}b^{5}(a+b)}{a^{3}+b^{3}}+\frac{b^{5}c^{5}(b+c)}{b^{3}+c^{3}}+\frac{a^{5}c^{5}(a+c)}{a^{3}+c^{3}} \geq\frac{3(a+b+c)}{(ab+ac+bc)}$