Conditional inequalities Let a,b,c be positive real numbers such that $abc=1$. Prove that $$\frac 1{a^3(b+c)}+\frac 1{b^3(c+a)}+\frac 1{c^3(a+b)} \ge \frac 32$$ We can derive the following inequalities from the given equality :-$$ab+bc+ca\ge3$$$$a+b+c\ge 3$$$$\frac 1a+\frac 1b+\frac 1c\ge3$$ None of them help to get a solution. If we simplify the given inequality, we get the following $$(a+b)(b+c)(\frac 2{b^3}-a-c)+(b+c)(c+a)(\frac 2{c^3}-a-b)+(c+a)(a+b)(\frac 2{a^3}-b-c) \ge 0$$ What do i do after this?

This is IMO 1995, Problem 2. 
$\sum \frac{1}{a^3(b+c)}=\sum\frac{1/a^2}{a(b+c)}\ge\sum \frac{(1/a+1/b+1/c)^2}{2(ab+bc+ca)}=\frac{ab+bc+ca}{2}\ge\frac{3\sqrt[3]{a^2b^2c^2}}{2}=\frac32.$
