An inequality with $\cos$ and triangle sides Here is the problem:

Let $ABC$ be a triangle with sides $a, b, c$. Show that $\dfrac{\cos A}{a^3}+\dfrac{\cos B}{b^3}+\dfrac{\cos C}{c^3}\geq\dfrac{3}{2abc}.$

Here's my attempt:
By the cosine formula, we have $\cos A = \dfrac{b^2+c^2-a^2}{2bc}$ etc, which the left hand side can be transformed into:
\begin{equation*}
\dfrac{a^2+b^2-c^2}{2abc^3}+\dfrac{a^2+c^2-b^2}{2ab^3c}+\dfrac{b^2+c^2-a^2}{2a^3bc}
\end{equation*}
And then I'm stuck. Can someone help me? Thanks.

You're almost done. If you factor out $\frac{1}{2abc}$ from the expression you obtained, you get
$$\frac{1}{2abc}\left(\frac{a^2+b^2-c^2}{c^2}+\frac{a^2+c^2-b^2}{b^2} + \frac{b^2+c^2-a^2}{a^2}\right).$$
So you just need to prove $\frac{a^2+b^2-c^2}{c^2}+\frac{a^2+c^2-b^2}{b^2} + \frac{b^2+c^2-a^2}{a^2}\ge 3$. Writing
\begin{align}&\frac{a^2+b^2-c^2}{c^2}+\frac{a^2+c^2-b^2}{b^2} + \frac{b^2+c^2-a^2}{a^2} \\
&= \left(\frac{a^2}{b^2}+\frac{b^2}{a^2}\right) + \left(\frac{a^2}{c^2}+\frac{c^2}{a^2}\right) + \left(\frac{b^2}{c^2}+\frac{c^2}{b^2}\right) - 3\end{align}
it is enough to show that each of the terms in the parentheses is at least $2$. Why is that true?
