## zmudz one year ago Let $$a, b, c,$$ and $$d$$ be the areas of the triangular faces of a tetrahedron, and let $$h_a, h_b, h_c,$$ and $$h_d$$ be the corresponding altitudes of the tetrahedron. If $$V$$ denotes the volume of the tetrahedron, find the maximum $$k$$ such that $$(a + b + c + d)(h_a + h_b + h_c + h_d) \geq kV.$$

1. thomas5267

So h are the altitude of the triangles which make up the tetrahedron?

2. zmudz

@thomas5267 from what I understand of the problem, yes

3. beginnersmind

|dw:1441666807027:dw| I think it's like this.

4. beginnersmind

I apologize for the terrible art.

5. beginnersmind

What are we looking at?

6. beginnersmind

@thomas5267 "So h are the altitude of the triangles which make up the tetrahedron?" No, they are the altitudes of the tetrahedron. As in V=(1/3)a*h_a

7. anonymous

|dw:1441667234670:dw|

8. beginnersmind

Anyway, I'd start with $$(a + b + c + d)(h_a + h_b + h_c + h_d) \geq kV.$$ $$(a + b + c + d)(\frac{V}{a} + \frac{V}{b} + \frac{V}{c} + \frac{V}{d} ) \geq 3kV.$$ $$(a + b + c + d)(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} ) \geq 3k.$$ Which seem like should be some sort of inequality between arithmetic and harmonic means.

9. anonymous

So $$h_n= \frac{V}{n}$$ where $$n$$ = face side?

10. beginnersmind

Anyway, we know that the arthmetic mean is greater than equal than the harmonic mean, so using some algebra we get that k is less then or equal to 1/3.

11. beginnersmind

@Jhannybean I used the volume formula for the tetrahedron $V = \frac{1}{3}a*h_a$ Where a is the area of a face and h_a is the corresponding altitude.

12. anonymous

Oh alright

13. beginnersmind

Correction: $V = \frac{1}{3}a*h_a$ $h_a =\frac{3V}{a}$ $(a + b + c + d)(h_a + h_b + h_c + h_d) \geq kV.$ $(a + b + c + d)(\frac{3V}{a} + \frac{3V}{b} + \frac{3V}{c} + \frac{3V}{d} ) \geq kV.$ $3(a + b + c + d)(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} ) \geq k.$ The inequality between arithmetic and harmonic means states that: $\frac{a_1+a_2+a_3+a_4}{4} \geq \frac{4}{\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4} }$ or $\frac{a_1+a_2+a_3+a_4}{16} \geq \frac{1}{\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4} }$ So $\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4} \geq \frac{1}{16}\frac{1}{a_1+a_2+a_3+a_4}$ At this point I'm actually confused. It seems like k has a minimum, not a maximum.

14. beginnersmind

E.g. substituting tells us that the LHS of the original expression is always greater or equal to 3/16, with equality happening if a=b=c=d

15. zmudz

oops, meant to say maximum. I'm pretty sure there is a maximum

16. beginnersmind

Actually, if you think about it geometrically, it's clear that there's no maximum. You can construct a tetrahedron like this: |dw:1441674391911:dw|

17. beginnersmind

|dw:1441674489410:dw| You can see that the areas of ABC (d) and ABD (c) are fairly large. Also, the altitude h_a is pretty long. So when you look at the product on the left hand side the terms c*h_a and d*h_a are large. However the volume V is quite small. So the constant k is large. Specifically we can make k arbitrarily large by "closing" the two large triangles.