Prove that for $a,b,c \in \mathbb{R}^+$ with $abc = 1$, $\frac{1}{a^3(b + c)} + \frac{1}{b^3(a + c)} + \frac{1}{c^3(a + b)} \ge \frac{3}{2}$

Question

I would like confirmation that I did this proof correctly. If I did, it would be a milestone in my mathematical journey as it would be my first IMO problem.

By Cauchy, we have that $$\left( \frac{1}{a^3(b + c)} + \frac{1}{b^3(a + c)} + \frac{1}{c^3(a + b)} \right) ( (b + c) + (a + c) + (a + b) ) \ge \left( \frac{1}{a^{3/2}} + \frac{1}{b^{3/2}} + \frac{1}{c^{3/2}} \right)^2.$$

By AM-GM, we have that $$\frac{1}{a^{3/2}} + \frac{1}{b^{3/2}} + \frac{1}{c^{3/2}} \ge 3\sqrt[3]{\frac{1}{a^{3/2}b^{3/2}c^{3/2}}} = 3\sqrt[3]{\frac{1}{(abc)^{3/2}}} = 3.$$

Then, again by AM-GM, we have that $a + b + c \ge 3\sqrt[3]{abc} = 3$, and so $(b + c) + (a + c) + (a + b) = 2(a + b + c) \ge 2\cdot3 = 6.$

Thus, we have that $$\left( \frac{1}{a^3(b + c)} + \frac{1}{b^3(a + c)} + \frac{1}{c^3(a + b)} \right) ( (b + c) + (a + c) + (a + b) ) \ge \left( \frac{1}{a^3(b + c)} + \frac{1}{b^3(a + c)} + \frac{1}{c^3(a + b)} \right)6 \ge 3^2 = 9,$$ and so $$\frac{1}{a^3(b + c)} + \frac{1}{b^3(a + c)} + \frac{1}{c^3(a + b)} \ge \frac 96 = \frac 32,$$ and we are done.

Answer 1

Your solution is incorrect (see my comment.)

One solution is indeed by Cauchy Schwartz:

$$\left( \frac{1}{a^3(b + c)} + \frac{1}{b^3(a + c)} + \frac{1}{c^3(a + b)} \right) ( a(b + c) + b(a + c) + c(a + b) )\ge(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2=\frac{(ab+bc+ca)^2}{(abc)^2}=(ab+bc+ca)^2$$

Thus, we have

$$\left( \frac{1}{a^3(b + c)} + \frac{1}{b^3(a + c)} + \frac{1}{c^3(a + b)} \right)\ge \frac{(ab+bc+ca)^2}{ ( a(b + c) + b(a + c) + c(a + b) )}=\frac{(ab+bc+ca)^2}{2(ab+bc+ca)}=\frac{ab+bc+ca}{2}$$

Since $abc=1$ so $ab+bc+ca\ge3\sqrt[3]{ab\times bc\times ca}=3\sqrt[3]{(abc)^2}=3$, and this can yield the result.