# Given that there exists a triangle whose sides are a,b,c. Then prove that there exists a triangle whose sqrta,sqrtb,sqrtc?

Jul 23, 2018

Given that there exists a triangle whose sides are a,b,c.

So we have following 3 inequalities satisfied.

• $a + b > c$

• $b + c > a$

• $c + a > b$

Considering the first one

$a + b > c$

$\implies {\left(\sqrt{a} + \sqrt{b}\right)}^{2} - 2 \sqrt{a b} > c$

$\implies {\left(\sqrt{a} + \sqrt{b}\right)}^{2} > c + 2 \sqrt{a b}$

$\implies \sqrt{a} + \sqrt{b} > \sqrt{c}$

Similarly from 2nd inequality we get

$\sqrt{b} + \sqrt{c} > \sqrt{a}$

And from 3rd inequality we get

$\sqrt{c} + \sqrt{a} > \sqrt{b}$

So we can say if there exists a triangle having sides $a , b \mathmr{and} c$ then there exists a triangle having sides $\sqrt{a} , \sqrt{b} \mathmr{and} \sqrt{c}$

Jul 24, 2018

My interest in the problem:

#### Explanation:

Choosing $a = 2 , b = 4 \mathmr{and} c = 3$, and using scale and compass

only, four conjoined $\triangle s D E F$,

with sides $\sqrt{a} = \sqrt{2} , \sqrt{b} = 2 \mathmr{and} \sqrt{c} = \sqrt{3}$ are constructed.

The graph shows one pair over the base $E F = \sqrt{2}$,

with vertices.

$D \left(\frac{1}{\sqrt{8}} , \pm \sqrt{\frac{23}{8}}\right) , E \left(0 , 0\right) \mathmr{and} F \left(\sqrt{2} , 0\right)$

There are three such pairs, and all have the central

common $\triangle D E F$

graph{(x^2+y^2-3)((x-sqrt2)^2+y^2-4)(x^2+y^2-0.01)((x-sqrt2)^2+y^2-0.01)((x-sqrt(1/8))^2+(y-sqrt((23)/8))^2-0.01)((x-sqrt(1/8))^2+(y+sqrt((23)/8))^2-0.01)=0[-4 4 -2 2]}