Vlados021 wrote: The tangents to the circumcircle of the acute triangle $ABC$, passing through $B$ and $C$, meet at point $F$. The points $M$, $L$, and $N$ are the feet of perpendiculars from the vertex $A$ to the lines $FB$, $FC$, and $BC$, respectively. Prove the inequality $AM+AL\ge 2AN$. (V. Karamzin) Quite trivial. $$b^2+c^2\geq 2bc \Longleftrightarrow \tfrac{c^2}{2R}+\tfrac{b^2}{2R}\geq \tfrac{2bc}{2R} \Longleftrightarrow AM+AL\geq 2\cdot AN$$

Perhaps, a more logical way is to express $AM, AL, AN$ Since, the triangle $ABC$ is acute WLOG $ b \ge c $ Thus $ \sin{b} \ge \sin{c}$ Now observe the inequality $AM + AL \ge 2 AN$ Which is equal to $ c \sin{c} + b \sin{b} \ge 2 AN = c \sin{b} + b \sin{c}$ Simplifying gives us $ \sin{b} \ge \sin{c}$ which is obviously true