Answer

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Why does this need to be bumped? Do you want a solution?

IAmTheHazard wrote: Why does this need to be bumped? Do you want a solution? Yes, I can't solve it.

$\delta(H, BC)h_A = bc\cos(B)\cos(C)$, $h_a = \frac{bc}{2R} = 2R\sin(B)\sin(C)$, $\delta(H, BC) = 2R\cos(B)\cos(C)$. $\frac{AH}{AA'} = \frac{h_A-\delta(H, BC) }{h_A+\delta(H, BC) } = \frac{\cos(A)}{\cos(B-C)}$ $\frac{\cos(A)}{\cos(B-C)} \geq \cos(A)$ Using Law of Cosines on the reflection of C over AH, we get that: $\frac{\cos(A)}{\cos(B-C)} = \frac{b^2+c^2-a^2}{b^2+c^2-(H_AB-H_AC)^2} $ We also know that $|H_AB-H_AC| = \frac{|b^2-c^2|}{a}$, so we have: $\frac{\cos(A)}{\cos(B-C)} = \frac{a^2b^2+a^2c^2-a^4}{a^2b^2+a^2c^2-b^4-c^4+2b^2c^2} $, cyclic sum is $(2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4)(\frac{1}{a^2b^2+a^2c^2-b^4-c^4+2b^2c^2}+\frac{1}{b^2c^2+c^2a^2-c^4-a^4+2c^2a^2}+\frac{1}{c^2a^2+c^2b^2-a^4-b^4+2a^2b^2}) - 3$ By Cauchy Schwarz, $\text{Sum} \geq \frac{9}{2}-3 = \frac{3}{2}$. $\frac{\cos(A)}{\cos(B-C)}=\frac{a^2b^2+a^2c^2-a^4}{a^2b^2+a^2c^2-b^4-c^4+2b^2c^2} < \frac{2a^2b^2+2a^2c^2-2a^4}{2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4}$ since $\triangle ABC$ is acute, so summing cyclically, $\text{Sum}<2$, we are done. $\blacksquare$

This has been already posted and I left my solution at the old post. My solution

@above Thank you two!