Question

show that the sum of the three altitudes of a triangle is less than the sum of the three sides of the triangle

Answer 1

Suppose we have a triangle ABC. Draw the altitude from vertex A, and let D be the point where it intersects the line BC.

Then ADB and ADC are both right triangles (possibly degenerate) with right angle D, and in a right triangle, the leg is always less than or equal to the hypotenuse. If necessary, you can see this fact using the Pythagorean theorem.

Anyway, this means that
AD <= AB
AD <= AC
so
2AD <= AB + AC

Applying this to each of the other altitudes:
2*(length of altitude with vertex B) <= AB + BC
2*(length of altitude with vertex C) <= AC + BC

Then when you add all these together,
2 * (sum of altitudes) <= 2 * (perimeter)
and therefore the sum of the altitudes is less than or equal to the perimeter.

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Note if you need "less than" and not "less than or equal to", just use the pythagorean theorem again in any of the above cases to show that equals only occurs when the triangle is fully degenerate (when it's a point).

Answer 2

Let there be a triangle ABC with its altitudes D, E, and F from vertices A, B and C respectively.

The altitudes form a right angle at their corresponding bases. Also, in a right triangle the hypotenuse is the longest side. Taking the right triangles formed by the altitudes and the sides as the hypotenuse, we observe that in each triangle, the side forms the longest side, i.e,

In triangle ABD, AB is the longest side

In triangle ACF, AC is the longest side

In triangle CBE, BC is the longest side

So, adding all the three, we get that the perimeter of a triangle is greater than the sum of its three altitudes.

Hope the answer is helpful !!

Good Day !!