jasperE3 wrote: Let $a,b,c$ be the sides and $\alpha,\beta,\gamma$ be the corresponding angles of a triangle. Prove the equality $$\left(\frac bc+\frac cb\right)\cos\alpha+\left(\frac ca+\frac ac\right)\cos\beta+\left(\frac ab+\frac ba\right)\cos\gamma=3.$$ Note that, from law of cosines we have $cos\alpha = \frac{b^2+c^2-a^2}{2bc}$. So, $$\sum_{cyc} (\frac{b}{c} + \frac{c}{b})cos\alpha = \sum_{cyc} \frac{a^2(b^2+c^2)^2 - a^4(b^2+c^2)}{2a^2b^2c^2} = \frac{6a^2b^2c^2}{2a^2b^2c^2} = 3$$

jasperE3 wrote: Let $a,b,c$ be the sides and $\alpha,\beta,\gamma$ be the corresponding angles of a triangle. Prove the equality $$\left(\frac bc+\frac cb\right)\cos\alpha+\left(\frac ca+\frac ac\right)\cos\beta+\left(\frac ab+\frac ba\right)\cos\gamma=3.$$ https://artofproblemsolving.com/community/c6h559886p3257934

jasperE3 wrote: Let $a,b,c$ be the sides and $\alpha,\beta,\gamma$ be the corresponding angles of a triangle. Prove the equality $$\left(\frac bc+\frac cb\right)\cos\alpha+\left(\frac ca+\frac ac\right)\cos\beta+\left(\frac ab+\frac ba\right)\cos\gamma=3.$$ It's well known that $b \cos \alpha + a \cos \beta = c$. To prove this without expanding, just note that in a triangle $ABC$ with altitude $CD$, then $b \cos \alpha$ is just $AD$ and $a \cos \beta$ is just $BD$, where $c = AB$. Therefore the above sum could be rearranged to \[ \sum_{cyc} \frac{b \cos \alpha + a \cos \beta}{c}= \sum_{cyc} 1 = 3 \]

a=b cos γ + c cos β b= a cos γ+ c cos α c= a cos β+ b cos α So if we arrange the LHS THen, (a cos β+ b cos α)/c + (a cos γ+ c cos α)/b + (b cos γ+ c cos β)/a = c/ c + b/b + a/a =1+1+1=3=RHS( proved) @ Krishijivi