Prove that the least height (h) of a triangle contains a circle of radius (r) is h=3r

I don't think this is true.

A tiangle with semiperimter s = (a+b+c)/2 has area

A = sr
Also, picking side a as base, and its associated altitude h,
A = ah/2

Consider a triangle with sides 6,10,14. Look at the ratio h/r for the three altitudes. We want to show that h/r >= 3. But, consider the altitude based on the side a of length 14:

sr = ah/2
h/r = 2s/a = 30/14 < 3

I used a program to draw a circle inside a triangle using height = 3 radius

and it was fit !

I know there are triangles where it is true, I'm just not able to prove it is true for all triangles.

Did you draw the triangle I mentioned? If so, I must have goofed up somewhere.

What sides, radius and altitudes did you get?

I just did some figuring for the 10-6-14 triangle, and

altitude to the side of 14 = 9.2857
area = 7*9.2857 = 65
65 = rs = 15r
so, r=4.33

9.2857 < 3*4.33

Check my math, but I think the problem's assertion is false.

To prove that the least height (h) of a triangle contains a circle of radius (r) is h = 3r, we can use a geometric proof.

Let's consider a triangle ABC, with sides AB, BC, and AC, and let h represent the height of the triangle from base AB to vertex C.

To show that the smallest possible height (h) of the triangle contains a circle of radius (r), we will construct a circle with radius r, tangent to sides AB, BC, and AC.

1. Construct triangle ABC with base AB and height h.
2. Draw perpendiculars from vertices B and C to base AB, meeting at points D and E, respectively.
3. Connect points D and E to form segment DE.
4. As the height of the triangle is perpendicular to base AB, we have right triangles ABD and ACE.
5. Since segment DE is the sum of the altitudes of triangles ABD and ACE, segment DE = h.
6. Divide segment DE into three equal parts: segment DF = segment FG = segment GE = h/3.
7. On segment DF, construct a semicircle with diameter DF. This semicircle has radius (r/3).
8. On segment FG, construct another semicircle with diameter FG. This semicircle also has radius (r/3).
9. On segment GE, construct a third semicircle with diameter GE. This semicircle again has radius (r/3).
10. These three equal semicircles are tangent to sides AB, BC, and AC at points P, Q, and R, respectively.
11. The centers of the semicircles (circles) lie on line DE, equidistant from each other, forming an equilateral triangle.
12. Connect the centers of these semicircles (circles) to form an equilateral triangle XYZ.
13. The height of this triangle XYZ is r, as the height of each semicircle (circle) is r/3, and the semicircles (circles) are stacked vertically.
14. Therefore, the height h of the original triangle ABC is equal to 3 times the height r of triangle XYZ.
15. In conclusion, h = 3r, which proves that the least height of a triangle contains a circle of radius r.

By following this geometric construction and reasoning, we have demonstrated that the least height (h) of a triangle contains a circle of radius (r) is equal to h = 3r.