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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.

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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?

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  • $\begingroup$ Um... sorry but I still don't know why the terms in the parentheses are at least 2. $\endgroup$ – blastzit Aug 7 '17 at 4:49
  • $\begingroup$ @blastzit The AM-GM inequality $\endgroup$ – iamwhoiam Aug 7 '17 at 4:56
  • $\begingroup$ @iamwhoiam ohhh thanks so much! $\endgroup$ – blastzit Aug 7 '17 at 4:57
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$$2abc\sum_{cyc}\frac {a^2+b^2-c^2}{2abc^3}=$$ $$=\sum_{cyc}(\frac {a^2}{c^2}+\frac {b^2}{c^2}-1)=$$ $$=-3+\sum_{cyc}(\frac {a^2}{b^2}+\frac {b^2}{a^2})=$$ $$=-3+\sum_{cyc}(2+(\frac {a}{b}-\frac {b}{a})^2\;)=$$ $$=+3+\sum_{cyc}(\frac {a}{b}-\frac {b}{a})^2\geq 3.$$

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