, 2023

Solution:

Let the circumcircle of $ABC$ meet the altitudes $AD$, $BE$, and $CF$ again at $I$, $J$, and $K$ respectively.

Lemma (9-point circle). $I$, $J$, $K$ are the reflections of $H$ across $BC$, $CA$, $AB$. Moreover, $D$, $E$, $F$, $X$, $Y$, $Z$ are the midpoints of $HI$, $HJ$, $HK$, $HA$, $HB$, $HC$.

Proof. Since $ABDE$ and $ABIC$ are cyclic, we see that
$$\angle EBD = \angle EAD = \angle CAI = \angle CBI.$$
Hence the lines $BI$ and $BH$ are reflections across $BC$. Similarly, $CH$ and $CI$ are reflections across $BC$, so $I$ is the reflection of $H$ across $BC$. The analogous claims for $J$ and $K$ follow. A $\times 2$ dilation from $H$ now establishes the result.

From this lemma, we get $AI = 2XD$, $BJ = 2EY$, and $CK = 2FZ$. Hence it is equivalent to showing that
$$\frac{AH}{2DX} + \frac{BH}{2EY} + \frac{CH}{2FZ} \geq \frac{3}{2}$$
which is in turn equivalent to
$$\frac{AH}{AI} + \frac{BH}{BJ} + \frac{CH}{CK} \geq \frac{3}{2}.$$
Let $a = JK$, $b = KI$ and $c = IJ$. Again by the lemma we find $AH = AK = AJ$, so by Ptolemy's theorem on $AKIJ$,
$$AJ \cdot KI + AK \cdot IJ = AI \cdot JK$$
Substituting and rearranging,
$$\begin{aligned} AH \cdot b + AH \cdot c & = AI \cdot a \\ AH \cdot (b + c) & = AI \cdot a \\ \frac{AH}{AI} & = \frac{a}{b + c}. \end{aligned}$$
Similarly,
$$\frac{BH}{BJ} = \frac{b}{c + a} \quad \text{and} \quad \frac{CH}{CK} = \frac{c}{a + b}$$
Plugging these back into $\left(^*\right)$, the desired inequality is now
$$\frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{a + b} \geq \frac{3}{2}.$$
This is known as Nesbitt's Inequality, which has many proofs. Below is one such proof.

Add 3 to both sides and rearrange:
$$\begin{aligned} \left(\frac{a}{b + c} + 1\right) + \left(\frac{b}{c + a} + 1\right) + \left(\frac{c}{a + b} + 1\right) & \geq \frac{3}{2} + 3 \\ \Longleftrightarrow \quad \frac{a + b + c}{b + c} + \frac{a + b + c}{c + a} + \frac{a + b + c}{a + b} & \geq \frac{9}{2} \\ \Longleftrightarrow \quad (a + b + c)\left(\frac{1}{b + c} + \frac{1}{c + a} + \frac{1}{a + b}\right) & \geq \frac{9}{2} \\ \Longleftrightarrow \quad & \frac{(b + c) + (c + a) + (a + b)}{3} \end{aligned}$$
which is true by the AM-HM inequality.