Indian National Mathematical Olympiad

Let $\angle C$ be the smallest angle, so that $CA \geq AB$ and $CB \geq AB$. In this case the altitude through $C$ is the longest one. Let the altitude through $C$ meet $AB$ in $D$ and let $H$ be the orthocentre of $ABC$. Let $CD$ extended meet the circumcircle of $ABC$ in $K$. We have $CD = h_{\max}$ so that the inequality to be proved is

$$AH + BH + CH \leq 2 CD$$

Using $CD = CH + HD$, this reduces to $AH + BH \leq CD + HD$. However, we observe that $AH = AK$, $BH = BK$ and $HD = DK$. (For example, $BH = BK$ and $DH = DK$ follow from the congruency of the right-angled triangles $DBK$ and $DBH$.)

Thus we need to prove that $AK + BK \leq CK$. Applying Ptolemy's theorem to the cyclic quadrilateral $BCAK$, we get

$$AB \cdot CK = AC \cdot BK + BC \cdot AK \geq AB \cdot BK + AB \cdot AK$$

This implies that $CK \geq AK + BK$, which is precisely what we are looking for.

There were other beautiful solutions given by students who participated in INMO-2009. We record them here.

1. Let $AD, BE, CF$ be the altitudes and $H$ be the orthocentre. Observe that

$$\frac{AH}{AD} = \frac{[AHB]}{[ADB]} = \frac{[AHC]}{[ADC]}$$

This gives

$$\frac{AH}{AD} = \frac{[AHB] + [AHC]}{[ADB] + [ADC]} = 1 - \frac{[BHC]}{[ABC]}$$

Similar expressions for the ratios $BH/BE$ and $CH/CF$ may be obtained. Adding, we get

$$\frac{AH}{AD} + \frac{BH}{BE} + \frac{CH}{CF} = 2$$

Suppose $AD$ is the largest altitude. We get

$$\frac{AH}{AD} + \frac{BH}{AD} + \frac{CH}{AD} \leq \frac{AH}{AD} + \frac{BH}{BE} + \frac{CH}{CF} = 2$$

This gives the result.

2. Let $O$ be the circumcentre and let $L, M, N$ be the midpoints of $BC, CA, AB$ respectively. Then we know that $AH = 2OL$, $BH = 2OM$ and $CH = 2ON$. As earlier, assume $AD$ is the largest altitude. Then $BC$ is the least side. We have

$$\begin{aligned} 4[ABC] &= 4[BOC] + 4[COA] + 4[AOB] \\ &= BC \times 2OL + CA \times 2OM + AB \times 2ON \\ &= BC \times AH + CA \times BH + AB \times CH \\ &\geq AB(AH + BH + CH) \end{aligned}$$

Thus

$$AH + BH + CH \leq \frac{4[ABC]}{AB} = 2AD$$

3. We make use of the fact that $AH = 2R \cos \angle A$, $BH = 2R \cos \angle B$, $CH = 2R \cos \angle C$ and $AD = 2R \sin \angle B \sin \angle C$, where $R$ is the circumradius of $ABC$. We are assuming that $AD$ is the largest altitude so that $\angle A$ is the least angle. Thus we have to prove that

$$\cos \angle A + \cos \angle B + \cos \angle C \leq 2 \sin \angle B \sin \angle C$$

under the assumption $\angle A \leq \angle B$ and $\angle A \leq \angle C$. On multiplying this by $2 \sin \angle A$, this is equivalent to

$$\begin{aligned} &2(\sin \angle A \cos \angle A + \sin \angle A \cos \angle B + \sin \angle A \cos \angle C) \\ &\leq 4 \sin \angle A \sin \angle B \sin \angle C = \sin 2A + \sin 2B + \sin 2C \end{aligned}$$

This is equivalent to

$$\cos \angle B (\sin \angle A - \sin \angle B) + \cos \angle C (\sin \angle A - \sin \angle C) \leq 0$$

Since $ABC$ is acute-angled and $A$ is the least angle, the result follows.