Question

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

Answer 1

Please keep drawing the figure as you follow below.
Let ABC be the triangle with sides BC, AC, AB opposite angles A, B, C being in length equal to a, b, c respectively.

Draw perpendiculars AD, BE and CF from A, B, C to opposite sides meeting sides BC, AC and AB at D, E, F respectively.

Now perpendicular AD² = AC² - CD² ==> AD² < AC² or  AD < AC
                                                                           or AD < b -----(1)
                  Also,  BE² = AB² - AE² ==> BE² < AB² or BE < AB
                                                                           or BE < c ------(2)
            Likewise , CF² = BC² - BF² ==> CF² < BC² or CF < BC 
                                                                           or CF < a ----(3)
 
Adding inequalities (1), (2), (3), AD + BE + CF < a + b + c

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 !!