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

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#

#=>(sqrta+sqrtb)^2-2sqrt(ab)>c#

#=>(sqrta+sqrtb)^2>c +2sqrt(ab)#

#=>sqrta+sqrtb>sqrtc#

Similarly from 2nd inequality we get

#sqrtb+sqrtc>sqrta#

And from 3rd inequality we get

#sqrtc+sqrta>sqrtb#

So we can say if there exists a triangle having sides #a,bandc# then there exists a triangle having sides #sqrta,sqrtbandsqrtc#

Choosing #a = 2, b = 4 and c = 3#, and using scale and compass

only, four conjoined #triangles DEF#,

with sides #sqrta = sqrt2, sqrt b = 2 and sqrt c = sqrt3# are constructed.

The graph shows one pair over the base #EF = sqrt2#,

with vertices.

#D ( 1/sqrt 8, +-sqrt (23/8) ), E ( 0, 0 ) and F ( sqrt2, 0 )#

There are three such pairs, and all have the central

common #triangle DEF#

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]}