In the interior of $\triangle ABC$ we take the arbitrary point $P$. Prove that the following inequality holds: $$\small c(\sin\angle CAP + \sin\angle CBP) + a(\sin\angle ABP +\sin\angle ACP) + b(\sin\angle BCP + \sin\angle BAP)\\ \le a + b + c$$
I tried to draw the distances from $P$ to the sides of the triangle and write the sines from the right angled triangles, but I couldn't get anything out of it. I saw that a case of equality is when the triangle is equilateral and $P$ is its center.
Can you help me?
Best Answer
To simplify derivation introduce the notation: $$ BC=a,\quad AC=b,\quad AB=c,\\ \measuredangle{CAB}=\alpha,\quad \measuredangle{ABC}=\beta,\quad \measuredangle{BCA}=\gamma;\\ \measuredangle{ABP}=\alpha_B,\quad \measuredangle{ACP}=\alpha_C,\\ \measuredangle{BAP}=\beta_A,\quad \measuredangle{BCP}=\beta_C,\\\ \measuredangle{CAP}=\gamma_A,\quad \measuredangle{CBP}=\gamma_B. $$ Then the initial inequality can be written as: $$\small \frac{a(\sin\alpha_B+\sin\alpha_C) +b(\sin\beta_A+\sin\beta_C)+c(\sin\gamma_A+\sin\gamma_B)}{a+b+c}\le1.\tag{*} $$ which can be proved as follows: $$\begin{align}\small &\scriptsize\frac{a(\sin\alpha_B+\sin\alpha_C) +b(\sin\beta_A+\sin\beta_C)+c(\sin\gamma_A+\sin\gamma_B)}{a+b+c}\\ &\scriptsize=\frac{\sin\alpha(\sin\alpha_B+\sin\alpha_C) +\sin\beta(\sin\beta_A+\sin\beta_C)+\sin\gamma(\sin\gamma_A+\sin\gamma_B)} {\sin\alpha+\sin\beta+\sin\gamma}\tag1\\ &\scriptsize=\frac{(\sin\beta\sin\beta_A+\sin\gamma\sin\gamma_A) +(\sin\alpha\sin\alpha_B+\sin\gamma\sin\gamma_B) +(\sin\alpha\sin\alpha_C+\sin\beta\sin\beta_C)} {\sin\alpha+\sin\beta+\sin\gamma}\tag2\\ &\scriptsize\le\frac{\sin\alpha+\sin\beta+\sin\gamma}{\sin\alpha+\sin\beta+\sin\gamma}=1\tag3. \end{align} $$
There may arise an interesting question when does the inequality $(\text{*})$ degenerate to equality?
It is easily seen from $(4)$ that this happens if and only if the following equations hold: $$ \begin {cases} \beta-\beta_A=\gamma-\gamma_A\\ \gamma-\gamma_B=\alpha-\alpha_B\\ \alpha-\alpha_C=\beta-\beta_C \end {cases}, $$ which together with $$ \begin {cases} \beta_A+\gamma_A=\alpha\\ \gamma_B+\alpha_B=\beta\\ \alpha_C+\beta_C=\gamma \end {cases} $$ implies: $$ \begin {cases} \beta_C=\gamma_B=\frac\pi2-\alpha\\ \gamma_A=\alpha_C=\frac\pi2-\beta\\ \alpha_B=\beta_A=\frac\pi2-\gamma \end {cases}. $$
Thus the equality holds if and only if the triangle $ABC$ is acute and $P$ is its circumcenter.
Note:
With the convention that the angle measures at a triangle vertex are signed and the direction towards the adjacent side is positive, the inequality (*) becomes valid for the whole plane. Correspondingly it degenerates to equality at the circumcenter of the triangle, regardless of its shape.