\frac 1{\sin^aA} \ge \frac{1}{\sin A}\Leftrightarrow \sin^{a-1}A \le 1 (luôn đúng)Ttự \Rightarrow \frac 1{\sin^aA}+\frac 1{\sin^bB}+\frac 1{\sin^cC} \ge \frac{1}{\sin A}+\frac 1{\sin B} +\frac 1{\sin C} (\star)\frac{1}{\sin A}+\frac 1{\sin B} +\frac 1{\sin C} \ge \frac 1{\cos \frac A2}+\frac 1{\cos \frac B2}+\frac 1{\cos \frac B2} (\star \star ) \frac 1{\cos \frac A2} \ge\frac 1{\sqrt[x]{\cos \frac A2}} \Leftrightarrow \cos^{x-1}\frac A2 \le 1 (luôn đúng)Ttự \Rightarrow \frac 1{\cos \frac A2}+\frac 1{\cos \frac B2}+\frac 1{\cos \frac C2} \ge \frac 1{\sqrt[x]{\cos \frac A2}}+\frac 1{\sqrt[y]{\cos \frac B2}}+\frac 1{\sqrt[z]{\cos \frac C2}} (\star \star \star)~~~~~~~~~~Từ (\star),(\star \star),(\star \star \star)\Rightarrow \frac 1{\sin^aA}+\frac 1{\sin^bB}+\frac 1{\sin^cC} \ge \frac 1{\sqrt[x]{\cos \frac A2}}+\frac 1{\sqrt[y]{\cos \frac B2}}+\frac 1{\sqrt[z]{\cos \frac C2}}Kết hợp dữ kiện đề bài suy ra dấu bằng xảy raTức là \begin{cases}x=y=z=1 \\a=b=c=1\\ \triangle ABC \quad \text{đều} \end{cases}\Rightarrow h_a=h_b=h_c,l_a=l_b=l_c\Rightarrow đpcm
\frac 1{\sin^aA} \ge \frac{1}{\sin A}\Leftrightarrow \sin^{a-1}A \le 1 (luôn đúng)Ttự
\Rightarrow \frac 1{\sin^aA}+\frac 1{\sin^bB}+\frac 1{\sin^cC} \ge \frac{1}{\sin A}+\frac 1{\sin B} +\frac 1{\sin C} (\star)$\
color{blue}{\frac{1}{\sin A}+\frac 1{\sin B} +\frac 1{\sin C} \ge \frac 1{\cos \frac A2}+\frac 1{\cos \frac B2}+\frac 1{\cos \frac B2} (\star \star )
} \frac 1{\cos \frac A2} \ge\frac 1{\sqrt[x]{\cos \frac A2}} \Leftrightarrow \cos^{x-1}\frac A2 \le 1
(luôn đúng)Ttự \Rightarrow \frac 1{\cos \frac A2}+\frac 1{\cos \frac B2}+\frac 1{\cos \frac C2} \ge \frac 1{\sqrt[x]{\cos \frac A2}}+\frac 1{\sqrt[y]{\cos \frac B2}}+\frac 1{\sqrt[z]{\cos \frac C2}} (\star \star \star)
~~~~~~~~~~Từ (\star),(\star \star),(\star \star \star)
\Rightarrow \frac 1{\sin^aA}+\frac 1{\sin^bB}+\frac 1{\sin^cC} \ge \frac 1{\sqrt[x]{\cos \frac A2}}+\frac 1{\sqrt[y]{\cos \frac B2}}+\frac 1{\sqrt[z]{\cos \frac C2}}$Kết hợp dữ kiện đề bài suy ra dấu bằng xảy raTức là $\begin{cases}x=y=z=1 \\a=b=c=1\\ \triangle ABC \quad \text{đều} \end{cases}\Rightarrow h_a=h_b=h_c,l_a=l_b=l_c$$\Rightarrow$ đpcm