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Let ABC be a triangle with $H$ as its the orthocenter and $AD,BE,CF$ as the three altitudes. Prove that $$\frac{AD}{HD}+\frac{BC}{HE}+\frac{CF}{HF}≥9$$

I substituted $AD$ for $AH+HD$, changing the inequality to $\frac{AH}{HD}+\frac{BH}{HE}+\frac{CH}{HF}≥6$ With a little trigonometry I converted this to $\frac{cosAcosB}{cosC}+\frac{cosBcosC}{cosA}+\frac{cosCcosA}{cosB}≥6$

I tried a few things here, but couldn't make much progress. For example, I took the common denominator and then applied AM GM on the numerator as all the terms were squares. This gave me $$\frac{(cosAcosB)^2+(cosBcosC)^2+(cosCcosA)^2}{cosAcosBcosC}≥3(cosAcosBcosC)^{1/3}$$ I think this is $>0$ but that can't be since we have to prove it to be ≥6.

Amadeus
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2 Answers2

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Hint: Express each individual ratio as the ratio as a ratio of areas.

There's a natural choice given the setup.
If you're stuck, state what you've tried, and why it doesn't work.

Then use $ ( a + b + c) ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} ) \geq 9$.

Calvin Lin
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I think you have made a mistake notice that $$\sum \frac{AH}{HD}=\sum \frac{|\cos A|}{|\cos B\cos C|}\ge 3\sqrt[3]{\frac {1}{|\cos A\cos B\cos C|}}$$

Now $${|\cos A\cos B\cos C|}\le 1/8\tag1$$ the last inequality is proved here

using $(1)$ can you finish it?

Albus Dumbledore
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  • yes, I had $\sum \frac{\cos A\cos B}{\cos C}$ instead of $\sum \frac{\cos A}{\cos B\cos C}$ thanks for your help! – Amadeus Nov 25 '20 at 07:07