<s>[INST] 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.
 [/INST] $$2abc\sum_{cyc}\frac {a^2+b^2-c^2}{2abc^3}=$$     $$=\sum_{cyc}(\frac {a^2}{c^2}+\frac {b^2}{c^2}-1)=$$    $$=-3+\sum_{cyc}(\frac {a^2}{b^2}+\frac {b^2}{a^2})=$$ $$=-3+\sum_{cyc}(2+(\frac {a}{b}-\frac {b}{a})^2\;)=$$ $$=+3+\sum_{cyc}(\frac {a}{b}-\frac {b}{a})^2\geq 3.$$
</s>