chechemuaen7

2022-03-16

An inequality with $\mathrm{cos}$ and triangle sides
Here is the problem:
Let ABC be a triangle with sides a,b,c. Show that $\frac{\mathrm{cos}A}{{a}^{3}}+\frac{\mathrm{cos}B}{{b}^{3}}+\frac{\mathrm{cos}C}{{c}^{3}}\ge \frac{3}{2abc}$

Roland Ramsey

By the cosine formula, we have $\mathrm{cos}A=\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}$ etc, which the left hand side can be transformed into:
$\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab{c}^{3}}+\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2a{b}^{3}c}+\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2{a}^{3}bc}$
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 $\frac{{a}^{2}+{b}^{2}-{c}^{2}}{{c}^{2}}+\frac{{a}^{2}+{c}^{2}-{b}^{2}}{{b}^{2}}+\frac{{b}^{2}+{c}^{2}-{a}^{2}}{{a}^{2}}\ge 3$. Writting
$\frac{{a}^{2}+{b}^{2}-{c}^{2}}{{c}^{2}}+\frac{{a}^{2}+{c}^{2}-{b}^{2}}{{b}^{2}}+\frac{{b}^{2}+{c}^{2}-{a}^{2}}{{a}^{2}}$
$=\left(\frac{{a}^{2}}{{b}^{2}}+\frac{{b}^{2}}{{a}^{2}}\right)+\left(\frac{{a}^{2}}{{c}^{2}}+\frac{{c}^{2}}{{a}^{2}}\right)+\left(\frac{{b}^{2}}{{c}^{2}}+\frac{{c}^{2}}{{b}^{2}}\right)-3$
it is enough to show that each of the terms in the parentheses is at least 2.

pintorreeqwf

$2abc\sum _{cyc}\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab{c}^{3}}=$
$=\sum _{cyc}\left(\frac{{a}^{2}}{{c}^{2}}+\frac{{b}^{2}}{{c}^{2}}-1\right)=$
$=-3+\sum _{cyc}\left(\frac{{a}^{2}}{{b}^{2}}+\frac{{b}^{2}}{{a}^{2}}\right)=$
$=-3+\sum _{cyc}\left(2+{\left(\frac{a}{b}-\frac{b}{a}\right)}^{2}\right)=$
$=+3+\sum _{cyc}{\left(\frac{a}{b}-\frac{b}{a}\right)}^{2}\ge 3$

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