If $abc=1$ where $a,b,c$ are positive real. Prove that ,$\frac{1}{a^3(b+c)} +\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)} \ge \frac 32$.
I tried to multiply the LHS by $abc$ to make the relation homogeneous but……
There is a $3$ in the RHS ,So I applied AM-GM and found that it is enough to show that
$(a+b)(b+c)(c+a) \le 8$ where $abc=1$.
I know it is a very well-known inequality but I couldn't answer it . Please help me.
Source: It is derived from a question came in any Olympiad question (I cannot remember, now).
It is a good problem for the application of AM-GM or, Cauchy-Schartz.Jensen's inequality can also help.
Best Answer
Let
$$a=\frac1x, \, b=\frac1y, \, c=\frac1z \implies xyz=1$$
$$\frac{1}{a^3(b+c)} +\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)} \ge \frac 32 \iff \frac{x^2}{y+z} +\frac{y^2}{x+z}+\frac{z^2}{x+y} \ge \frac 32$$
by Jensen's Inequality with $f(x)=x^{-1}$ convex we have that
$$\frac{\frac{x^2}{y+z} +\frac{y^2}{x+z}+\frac{z^2}{x+y}}{x+y+z}=\frac{xf\left(\frac{y+z}x\right) +yf\left(\frac{x+z}y\right) +zf\left(\frac{x+y}z\right)}{x+y+z}\ge f\left(\frac{(y+z)+(x+z)+(x+y)}{x+y+z}\right)=\frac12 $$
and by AM-GM $x+y+z\ge 3\sqrt[3]{xyz}=3$.