Inequality in 3 variables with a constraint condition To prove : a(b+c)/bc + b(c+a)/ca + c(a+b)/ab > 2/(ab+bc+ca) where a+b+c=1 and a,b,c are positive real numbers
Here's my way : Add 1 to each summand in the LHS and subtract 3 (=1+1+1) and after some algebraic manipulation I got (ab+bc+ac)/abc > 2/(ab+bc+ac) +3 which again after some algebra reduces to a²b²+b²c²+a²c² +2abc(a+b+c) > 2+3abc(a+b+c)=> a²b²+b²c²+a²c² > 2+abc . Now, this obviously true for positive a,b,c but I can't prove it rigorously this. 
 A: If we arrange
$$
\frac{ab+bc+ac}{abc} \ge \frac{2}{ab+bc+ac} +3
$$ then we have
$$
(ab+bc+ac)^2 =\sum_{\text{cyc}}a^2b^2 +2abc\ge abc(2+3(ab+bc+ac))=2abc +3\sum_{\text{cyc}}a^2b^2c.
$$ We need to show
$$
\sum_{\text{cyc}}a^2b^2\ge 3\sum_{\text{cyc}}a^2b^2c.
$$ This is equivalent to
$$
\sum_{\text{cyc}}a^2b^2(a+b+c)\ge 3\sum_{\text{cyc}}a^2b^2c.
$$ By rearrangement, we have
$$
\sum_{\text{cyc}}a^3b^2\ge \sum_{\text{cyc}}a^2b^2c,
$$and
$$
\sum_{\text{cyc}}a^2b^3\ge \sum_{\text{cyc}}a^2b^2c.
$$ This proves the inequality.
Another approach: We need to show
$$
\sum_{\text{cyc}}\frac{a}{b}+\sum_{\text{cyc}}\frac{a}{c}\ge \frac{2}{ab+bc+ca}.
$$ By C-S, we have
$$
\sum_{\text{cyc}}\frac{a}{b}\sum_{\text{cyc}}ab\ge \left(\sum_{\text{cyc}}a\right)^2 =1,
$$ and
$$
\sum_{\text{cyc}}\frac{a}{c}\sum_{\text{cyc}}ac\ge \left(\sum_{\text{cyc}}a\right)^2 =1.
$$
A: Because by C-S
$$(ab+ac+bc)\sum_{cyc}\frac{a(b+c)}{bc}=\sum_{cyc}ac\sum_{cyc}\frac{a}{c}+\sum_{cyc}ab\sum_{cyc}\frac{a}{b}\geq$$
$$\geq(a+b+c)^2+(a+b+c)^2=2.$$
