Prove that the sum of distances of the orthocenter from the sides of an acute triangle is less than or equal to \(3r\), where \(r\) is the inradius of the same triangle.

If \(D, E, F\) are the feet of the altitudes from \(A, B, C\), respectively, to the opposite sides, and \( H \) the orthocentre of \( \triangle ABC \),
\[ 2\Delta = a\cdot HD + b\cdot HE + c\cdot HF = r(a + b + c) \]

\( \Delta \) is the area of \( \triangle ABC \), and \( r \) is its inradius.

\[ r = \frac{a\cdot HD + b\cdot HE + c\cdot HF}{a + b + c} \]

Note that \((a, b, c)\) and \((HD, HE, HF)\) are similarily sorted. (As \(HD = 2R \cos B \cos C \) and so on.)
Then, we can apply Chebyshev's inequality of the means, (basically the Rearrangement Inequality added many times)

\[ \frac{a\cdot HD + b\cdot HE + c\cdot HF}{3} \geq \frac{a + b + c}{3} \cdot \frac{HD + HE + HF}{3} \]
This gives,
\[ r = \frac{a\cdot HD + b\cdot HE + c\cdot HF}{a + b + c} \geq \frac{HD + HE + HF}{3} \]
as required.

Equality iff \( a = b = c \iff HD = HE = HF \) as is apparent from the Rearrangement Inequality condition.

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## Comments

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TopNewestIf \(D, E, F\) are the feet of the altitudes from \(A, B, C\), respectively, to the opposite sides, and \( H \) the orthocentre of \( \triangle ABC \), \[ 2\Delta = a\cdot HD + b\cdot HE + c\cdot HF = r(a + b + c) \]

\( \Delta \) is the area of \( \triangle ABC \), and \( r \) is its inradius.

\[ r = \frac{a\cdot HD + b\cdot HE + c\cdot HF}{a + b + c} \]

Note that \((a, b, c)\) and \((HD, HE, HF)\) are similarily sorted. (As \(HD = 2R \cos B \cos C \) and so on.)

Then, we can apply Chebyshev's inequality of the means, (basically the Rearrangement Inequality added many times)

\[ \frac{a\cdot HD + b\cdot HE + c\cdot HF}{3} \geq \frac{a + b + c}{3} \cdot \frac{HD + HE + HF}{3} \] This gives, \[ r = \frac{a\cdot HD + b\cdot HE + c\cdot HF}{a + b + c} \geq \frac{HD + HE + HF}{3} \] as required.

Equality iff \( a = b = c \iff HD = HE = HF \) as is apparent from the Rearrangement Inequality condition.

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