Jovany Clayton

Answered

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?

$a\mathrm{cos}A+b\mathrm{cos}B+c\mathrm{cos}C\le \frac{a+b+c}{2}$

How to initiate this question?

Answer & Explanation

kawiarkahh

Expert

2022-07-01Added 15 answers

If $a\ge b\ge c$ then $\alpha \ge \beta \ge \gamma $ and since cos is a decreasing function on $[0,\pi ]$ , 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:

$\sum _{cyc}a\mathrm{cos}\alpha \le \frac{1}{3}(a+b+c)(\mathrm{cos}\alpha +\mathrm{cos}\beta +\mathrm{cos}\gamma )\le \frac{a+b+c}{2}.$

Done!

$\mathrm{cos}\alpha \le \mathrm{cos}\beta \le \mathrm{cos}\gamma .$

Thus, since

$\mathrm{cos}\alpha +\mathrm{cos}\beta +\mathrm{cos}\gamma =1+\frac{r}{R}\le \frac{3}{2},$

by Chebyshov we obtain:

$\sum _{cyc}a\mathrm{cos}\alpha \le \frac{1}{3}(a+b+c)(\mathrm{cos}\alpha +\mathrm{cos}\beta +\mathrm{cos}\gamma )\le \frac{a+b+c}{2}.$

Done!

Esmeralda Lane

Expert

2022-07-02Added 7 answers

We need to prove that

$\sum _{cyc}\frac{a({b}^{2}+{c}^{2}-{a}^{2})}{2bc}\le \frac{a+b+c}{2}$

or

$\sum _{cyc}{a}^{2}({b}^{2}+{c}^{2}-{a}^{2})\le \sum _{cyc}{a}^{2}bc$

or

$\sum _{cyc}({a}^{4}-2{a}^{2}{b}^{2}+{a}^{2}bc)\ge 0$

or

$\sum _{cyc}({a}^{4}-{a}^{3}b-{a}^{3}c+{a}^{2}bc)+\sum _{cyc}({a}^{3}b+{a}^{3}c-2{a}^{2}{b}^{2})\ge 0,$

which is Schur and $\sum _{cyc}ab(a-b{)}^{2}\ge 0.$

Done!

$\sum _{cyc}\frac{a({b}^{2}+{c}^{2}-{a}^{2})}{2bc}\le \frac{a+b+c}{2}$

or

$\sum _{cyc}{a}^{2}({b}^{2}+{c}^{2}-{a}^{2})\le \sum _{cyc}{a}^{2}bc$

or

$\sum _{cyc}({a}^{4}-2{a}^{2}{b}^{2}+{a}^{2}bc)\ge 0$

or

$\sum _{cyc}({a}^{4}-{a}^{3}b-{a}^{3}c+{a}^{2}bc)+\sum _{cyc}({a}^{3}b+{a}^{3}c-2{a}^{2}{b}^{2})\ge 0,$

which is Schur and $\sum _{cyc}ab(a-b{)}^{2}\ge 0.$

Done!

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