Jovany Clayton

2022-06-30

For any triangle ABC with sides a,b,c prove that
$a\mathrm{cos}A+b\mathrm{cos}B+c\mathrm{cos}C\le \frac{a+b+c}{2}$
How to initiate this question?

kawiarkahh

Expert

If $a\ge b\ge c$ then $\alpha \ge \beta \ge \gamma$ and since cos is a decreasing function on $\left[0,\pi \right]$ , we obtain:
$\mathrm{cos}\alpha \le \mathrm{cos}\beta \le \mathrm{cos}\gamma .$
Thus, since
$\mathrm{cos}\alpha +\mathrm{cos}\beta +\mathrm{cos}\gamma =1+\frac{r}{R}\le \frac{3}{2},$
by Chebyshov we obtain:
$\sum _{cyc}a\mathrm{cos}\alpha \le \frac{1}{3}\left(a+b+c\right)\left(\mathrm{cos}\alpha +\mathrm{cos}\beta +\mathrm{cos}\gamma \right)\le \frac{a+b+c}{2}.$
Done!

Esmeralda Lane

Expert

We need to prove that
$\sum _{cyc}\frac{a\left({b}^{2}+{c}^{2}-{a}^{2}\right)}{2bc}\le \frac{a+b+c}{2}$
or
$\sum _{cyc}{a}^{2}\left({b}^{2}+{c}^{2}-{a}^{2}\right)\le \sum _{cyc}{a}^{2}bc$
or
$\sum _{cyc}\left({a}^{4}-2{a}^{2}{b}^{2}+{a}^{2}bc\right)\ge 0$
or
$\sum _{cyc}\left({a}^{4}-{a}^{3}b-{a}^{3}c+{a}^{2}bc\right)+\sum _{cyc}\left({a}^{3}b+{a}^{3}c-2{a}^{2}{b}^{2}\right)\ge 0,$
which is Schur and $\sum _{cyc}ab\left(a-b{\right)}^{2}\ge 0.$
Done!

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