Prove that semi-perimeter $\geq a \cos A+b \cos B+c \cos C $

Question

For any triangle $ABC$ with sides $a,b,c$ prove that

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

How to initiate this question? Could someone give me some hint?

Answer 1

We need to prove that $$\sum_{cyc}\frac{a(b^2+c^2-a^2)}{2bc}\leq\frac{a+b+c}{2}$$ or $$\sum_{cyc}a^2(b^2+c^2-a^2)\leq\sum_{cyc}a^2bc$$ or $$\sum_{cyc}(a^4-2a^2b^2+a^2bc)\geq0$$ or $$\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)+\sum_{cyc}(a^3b+a^3c-2a^2b^2)\geq0,$$ which is Schur and $\sum\limits_{cyc}ab(a-b)^2\geq0.$

Done!

Answer 2

Let $R$ denote the radius of $\triangle ABC$'s circumcircle, we have $$a = 2R\sin A,b=2R\sin B, c=2R\sin C.$$ We then need to show that $$ 2\sin A\cos A + 2\sin B \cos B + 2 \sin C\cos C \le \sin A + \sin B +\sin C.$$

The left hand side can be written as $$ \begin{aligned}\sin 2A + \sin 2B + 2 \sin C\cos C &= 2\sin(A+B)\cos(A-B) - 2\sin C \cos (A+B)\\ &= 2\sin C (\cos(A-B)-\cos(A+B))\\ &= 4 \sin A \sin B \sin C. \end{aligned}$$

Thus we need to show that $$4\sin A \sin B\sin C \le \sin A + \sin B +\sin C,$$

which is obvious from the fact that $$\sin A\sin B \sin C \le \frac{3\sqrt 3}{8}.$$

Answer 3

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