1) Cho $a,b,c$ dương $ab+bc+ac=abc$ . C/m :
$\sum \frac{1}{a+2b+3c}<\frac{3}{16}$
2) Cho $a,b,c$ dương. $a+b+c=1$. C/m : $\frac{3}{ab+bc+ac}+\frac{2}{a^2+b^2+c^2}>14$
3) Tìm Min biết $a,b$ dương $a+b \ge 1$ . $B=\frac{1}{a^2+ab}+\frac{1}{b^2+ab}$
2)-Ta có: \[ab + bc + ca \le \frac{{{{(a + b + c)}^2}}}{3} = \frac{1}{3} = > \frac{1}{{ab + bc + ca}} \ge 3(1).\]
-Và có: \[\frac{1}{{ab + bc + ca}} + \frac{1}{{ab + bc + ca}} + \frac{2}{{{a^2} + {b^2} + {c^2}}} \ge \frac{{{{(1 + 1 + \sqrt 2 )}^2}}}{{{{(a + b + c)}^2}}} = 6 + 4\sqrt 2 (2).\]
-Từ (1);(2) => \[\frac{3}{{ab + bc + ca}} + \frac{2}{{{a^2} + {b^2} + {c^2}}}\]\[ \ge 3 + 6 + 4\sqrt 2 > 14\].
=> đpcm.