MHB Triangle Inequality: $a^4-1, a^4+a^3+2a^2+a+1, 2a^3+a^2+2a+1$

[anemone]

Show that for all $a>1$, there is a triangle with sides $a^4-1$, $a^4+a^3+2a^2+a+1$, and $2a^3+a^2+2a+1$.

 

[kaliprasad]

Show that for all $a>1$, there is a triangle with sides $a^4-1$, $a^4+a^3+2a^2+a+1$, and $2a^3+a^2+2a+1$.

Spoiler

Let the Lengths of sides of triangle be
$x = a^4 – 1$
$y = a^4 + a^3 + 2a^2 + a +1$
$z= 2a^3 + a^2 + 2a + 1$

clearly x < y
now we need to see if y < z or y = z or y > z
$y – z = a^4 – a^3 + a^2 – a = a^3(a-1) +a (a-1) >0$
so y – z >0
so y is the longer side,
now if we prove that x + z > y then we are through
$x + z – y = a^3 – a^2 + a – 1 = (a^2+1)(a-1) > 0$

hence proved

 

[DreamWeaver]

Great problem, Anemone! :D

And a very elegant solution, Kaliprasad! :D

 

[anemone]

Great problem, Anemone! :D

It feels quite nice to receive such a compliment from time to time at MHB for my posting of the challenge problem(s)!:p(Sun)

 

[Albert1]

Show that for all $a>1$, there is a triangle with sides $a^4-1$, let :$a^4+a^3+2a^2+a+1$, and $2a^3+a^2+2a+1$.

Spoiler

let:
$x=a^4-1=(a^2+1)(a^2-1)$
$y=a^4+a^3+2a^2+a+1=(a^2+a+1)(a^2+1)$
and
$z=2a^3+a^2+2a+1=(a^2+1)(2a+1)$
if $x,y,z $ can form a triangle ,then :
$xx=a^2-1$
$yy=a^2+a+1$
$zz=2a+1$
can also form a new but smaller triangle (by shrinking $a^2+1$ fold)
again $yy$ is the longest
now we must prove $xx+zz>yy$ if $a>1$
but $xx+zz-yy=a^2-1+2a+1-a^2-a-1=a-1>0(\,\, if \,\, a>1)$
and the proof is done