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$\frac 1{\sin^aA} \ge \frac{1}{\sin A}\Leftrightarrow \sin^{a-1}A \le 1$ (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