# An Inequality in Triangle, VI

### Proof 1

$\displaystyle\frac{1}{2}ah_a=\frac{1}{2}bh_b=\frac{1}{2}ch_c.\;$ Hence, $h_a=bck,\;$ $h_b=ack,\;$ $c=abk,$ $k\gt 0.\;$ A substitution gives $\displaystyle\frac{c}{a+b}\lt 1\lt\frac{c}{|a-b|},\;$ which is equivalent to $|a-b|\lt c\lt a+b.$

### Proof 2

$ah_a=bh_b=ch_c=2E,\;$ so that $\displaystyle h_a=\frac{2E}{a},\;$ $\displaystyle h_b=\frac{2E}{b},\;$ $\displaystyle h_c=\frac{2E}{c}.\;$ Thus, $\displaystyle\frac{h_a\cdot h_b}{h_a+h_b}\lt h_c\lt \frac{h_a\cdot h_b}{|h_a-h_b|}\;$ is equivalent to

$\displaystyle\frac{\displaystyle \frac{2E}{a}\cdot \frac{2E}{b}}{\displaystyle \frac{2E}{a}+\frac{2E}{b}}\lt\frac{2E}{c}\lt \frac{\displaystyle \frac{2E}{a}\cdot \frac{2E}{b}}{\left|\displaystyle \frac{2E}{a}-\frac{2E}{b}\right|},$

which simplifies to

$\displaystyle\frac{\displaystyle \frac{1}{a}\cdot \frac{1}{b}}{\displaystyle \frac{1}{a}+\frac{1}{b}}\lt\frac{2E}{c}\lt \frac{\displaystyle \frac{1}{a}\cdot \frac{1}{b}}{\left|\displaystyle \frac{1}{a}-\frac{1}{b}\right|}.$

Simplifying further, this is equivalent to $\displaystyle\frac{1}{a+b}\lt\frac{1}{c}\lt\frac{1}{|a-b|}\;$ and finally to $|a-b|\lt c\lt a+b.$

### Proof 3

$ah_a=bh_b=ch_c=2E,\;$ so that $\displaystyle h_a=\frac{2E}{a},\;$ $\displaystyle h_b=\frac{2E}{b},\;$ $\displaystyle h_c=\frac{2E}{c}.\;$ Thus, $\displaystyle\frac{h_a\cdot h_b}{h_a+h_b}\lt h_c\lt \frac{h_a\cdot h_b}{|h_a-h_b|}\;$ is equivalent to

$\displaystyle\frac{\displaystyle \frac{2E}{a}\cdot \frac{2E}{b}}{\displaystyle \frac{2E}{a}+\frac{2E}{b}}\lt\frac{2E}{c}\lt \frac{\displaystyle \frac{2E}{a}\cdot \frac{2E}{b}}{\left|\displaystyle \frac{2E}{a}-\frac{2E}{b}\right|},$

which simplifies to

$\displaystyle\frac{\displaystyle \frac{1}{ab}}{\displaystyle \frac{a+b}{ab}}\lt\frac{2E}{c}\lt \frac{\displaystyle \frac{1}{ab}}{\displaystyle \frac{|a-b|}{ab}}.$

Simplifying further, this is equivalent to $\displaystyle\frac{1}{a+b}\lt\frac{1}{c}\lt\frac{1}{|a-b|}\;$ and finally to $|a-b|\lt c\lt a+b.$

### Proof 4

We know that if $\displaystyle x=\frac{1}{h_a},\;$ $\displaystyle y=\frac{1}{h_b},\;$ and $\displaystyle z=\frac{1}{h_c},\;$ then the triangle with sides $x,y,z\;$ is similar to $\Delta ABC.\;$ We replace by $x,y,z\;$ in our inequality and get the required result.

### Proof 5

The required inequality is implied by $\displaystyle\left|\frac{1}{h_a}-\frac{1}{h_b}\right|\lt\frac{1}{h_c}\lt\frac{1}{h_a}+\frac{1}{h_b}\;$ which is also true for isosceles triangles because, due to $ah_a=bh_b=ch_c,\;$ it is equivalent to the triangle inequality $|a-b|\lt c\lt a+b.$

### Acknowledgment

The inequality has been kindly posted at the CutTheKnotMath facebook page by Dorin Marghidanu along with four proofs. Proof 1 is by Kunihiko Chikaya; Proof 2 by Vaggelis Stamatiadis; Proof 3 by Dorin Marghidanu; Proof 4 by Leo Giugiuc.