$\frac{AH}{DX} = \frac{2AH}{AH_A} = \frac{2H_BH_C}{H_A H_B + H_AH_B}$ and cyclic variants. Result easily follows

eduD_looC wrote: An acute triangle is a triangle that has all angles less than $90^{\circ}$ ($90^{\circ}$ is a Right Angle). Let $ABC$ be an acute triangle with altitudes $AD$, $BE$, and $CF$ meeting at $H$. The circle passing through points $D$, $E$, and $F$ meets $AD$, $BE$, and $CF$ again at $X$, $Y$, and $Z$ respectively. Prove the following inequality: $$\frac{AH}{BX}+\frac{BH}{EY}+\frac{CH}{FZ} \geq 3.$$ Does the problem say $DX$ or $BX$ in the denominator? (The post above uses $DX$)

better have been $DX$

$\frac{AH}{AD}+\frac{BH}{BE}+\frac{CH}{CF}=2$ by areas, let $a=\frac{AH}{AD}$ etc. 9 pt circle gives $AX=\frac12 AH$, $\frac{AH}{DX}=\frac{1}{\frac{AD-AX}{AH}}=\frac{1}{\frac1a - \frac12}=\frac{2a}{2-a}=\frac{2a}{b+c}$ so we need $\sum_{\text{cyc}} \frac{2a}{b+c}\ge 3$ which is well known

Since $X$ is midpoint of $AH$, we have to prove $\sum \frac{AX} {DX} \geq \frac {3}{2}$. By angle chasing and law of sines in $\triangle XED$ we have $\frac{AX} {DX}=\frac{EX} {DX}=\frac{\cos \alpha} {\cos (\beta-\gamma)}=\frac{-\cos (\beta+\gamma)} {\cos(\beta-\gamma)} =\frac {\sin \beta. \sin \gamma-\cos \beta. \cos \gamma} {\sin \beta. \sin \gamma+\cos \beta. \cos \gamma}=\frac{\tan \beta. \tan \gamma-1}{\tan \beta. \tan \gamma+1}$. Using $\tan \alpha. \tan \beta. \tan \gamma=\tan \alpha+\tan \beta +\tan \gamma$, the last expression equals $\frac {\tan \beta+\tan \gamma} {2\tan \alpha+\tan \beta +\tan \gamma}$, so if we let $x=\tan \beta+\tan \gamma, y=\tan \alpha+\tan \gamma, z=\tan \alpha+\tan \beta$, then the inequality transforms to $\sum \frac {x} {y+z} \geq \frac{3}{2}$, which is Nesbitt.

Firstly, since $(DEF)$ is $(ABC)$ under a $\frac12$-homothety at $H$, we have that $X,Y,Z$ are the midpoints of $AH,BH,CH$. Then, note that \[\frac{AH}{DX} = \frac{AH \cdot BC }{DX\cdot BC} = \frac{2([AHB] + [AHC])}{2[BHC] + [AHB] + [AHC]}\]If we let $c = [AHB], b = [AHC], a=[BHC],$ we get \[\frac{AH}{DX} = 2\frac{b+c}{2a+b+c}\]Thus, it suffices to show \[\sum \frac{b+c}{2a+b+c} \geq \frac32\]Homogenize so that $a+b+c=1$, so we have \[\sum \frac{b+c}{2a+b+c} = \sum \frac{1-a}{1+a}\]Note that if $f(x) = \frac{1-x}{1+x}$, then $f'(x) = -\frac{2}{(x+1)^2}$, which is clearly increasing, so $f''(x)>0$ and this function is convex, so $ \sum \frac{1-a}{1+a}$ is minimized when $a=b=c = \frac13$, recovering $\frac32$ and we're done. $\blacksquare$.

Sketch: Set up a three variable inequality where the variables where the variables sum to $1$, derived by an argument involving areas. Then Jensen's pretty much finishes

It is well known that $X$,$Y$ , $Z$ are midpoints of $AH, BH,CH$ respectively. Also by areas we have $\frac{AH}{AD}+\frac{BH}{BE}+\frac{CH}{CF}=2$ so set $a=\frac{AH}{AD}, b=\frac{BH}{BE}, c=\frac{CH}{CF}$ we have $a+b+c=2$ Also we have $\frac{AH}{DX}=\frac{AH}{AD-AX}=\frac{AH}{AD-\frac{AH}{2}}=\frac{1}{\frac{1}{a}-\frac{1}{2}}=\frac{2a}{2-a}$ so we have $$\frac{AH}{DX}+\frac{BH}{EY}+\frac{CH}{FZ} =\frac{2a}{2-a}+\frac{2b}{2-b}+\frac{2c}{2-c}$$ we have this to be $2\left( -3+\sum_{a,b,c} \frac{2}{2-a}\right)$ and from titu's lemma we have $\sum_{a,b,c} \frac{1}{2-a} \geqslant \frac{9}{4}$ so we have $$\frac{AH}{DX}+\frac{BH}{EY}+\frac{CH}{FZ} \geq 3.$$as desired. $\blacksquare$

$\frac{AH}{DX}+\frac{BH}{EY}+\frac{CH}{FZ} \geq 3 \iff (\frac{XH}{DX}+\frac{YH}{EY}+\frac{ZH}{FZ})\geq \frac{3}{2}$ So now the problem is equivalent to the following problem Equiv Prob :- Let $ABC$ be a triangle with incenter $I$.$AI$ meet circumcircle again at $X$. Define $Y,Z$ similarly.Prove that $(\frac{XI}{AX}+\frac{YI}{BY}+\frac{ZI}{CZ})\geq \frac{3}{2}$. Proof Let $BC,CA,AB$ be denoted by $a,b,c$.By incenter excenter lemma ($XB=XI=XC$) and ptolemy's theorem we get $XI \cdot (b+c)=AX \cdot a \implies \frac{XI}{AX}=\frac{a}{b+c}$. Hence $(\frac{XI}{AX}+\frac{YI}{BY}+\frac{ZI}{CZ})=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{3}{2}$ follows by nesbitt's inequality.

Solution using ptolemy theorem Let $A'=AH \cap (ABC)$ and define $B',C'$ similarly Note $\triangle A'B'C' \sim \triangle DEF$ with ratio $2$. As $X,Y,Z$ are midpoint of $AH,BH,CH,$ (As they lie on Nine point circle) we need to prove $$\frac{AH}{AA'}+\frac{BH}{BB'}+\frac{CH}{CC'} \geq \frac{3}{2}$$ Note $HA=HC'=HB'$ (from reflection property of orthocenter). Now we use ptolemy theorem $$HC'.B'A' + HB'.A'C' = AA'.C'B'$$ from $A'B' = 2 DE$ and $ A'C' = 2 DF$ we get $$\frac{AH}{AA'} = \frac{EF}{DE+EF}$$ similarly doing for all give us $$ \frac{AH}{AA'}+\frac{BH}{BB'}+\frac{CH}{CC'} = \frac{EF}{DE+DF} + \frac{DF}{DE+EF} + \frac{DE}{DF+EF}$$ By Nesbitt's Inequality we are done $\blacksquare$