Triangle Challenge: Prove $p^4+q^4+r^4-2p^2q^2-2q^2r^2-2r^2p^2<0$

[anemone]

Prove that $p^4+q^4+r^4-2p^2q^2-2q^2r^2-2r^2p^2<0$ for $p,\,q,\,r$ are the sides of a triangle.

 

[Euge]

Prove that $p^4+q^4+r^4-2p^2q^2-2q^2r^2-2r^2p^2<0$ for $p,\,q,\,r$ are the sides of a triangle.

Spoiler

By the triangle inequality, $|p - q| < r < p + q$. Thus $(p - q)^2 < r^2 < (p + q)^2$; Using this, we find

$\displaystyle p^4 + q^4 + r^4 - 2p^2q^2 - 2q^2r^2 - 2r^2p^2$

$\displaystyle = (p^2 + q^2 - r^2)^2 - 4p^2q^2$

$\displaystyle = (p^2 + q^2 - r^2 - 2pq)(p^2 + q^2 - r^2 + 2pq)$

$\displaystyle = [(p - q)^2 - r^2][(p + q)^2 - r^2]$

$\displaystyle < 0$.

 

[anemone]

Spoiler

By the triangle inequality, $|p - q| < r < p + q$. Thus $(p - q)^2 < r^2 < (p + q)^2$; Using this, we find

$\displaystyle p^4 + q^4 + r^4 - 2p^2q^2 - 2q^2r^2 - 2r^2p^2$

$\displaystyle = (p^2 + q^2 - r^2)^2 - 4p^2q^2$

$\displaystyle = (p^2 + q^2 - r^2 - 2pq)(p^2 + q^2 - r^2 + 2pq)$

$\displaystyle = [(p - q)^2 - r^2][(p + q)^2 - r^2]$

$\displaystyle < 0$.

Hey Euge, you're so tremendously great at factoring to simplify any given math expressions and I admire all those heuristic skills that you posses! So I tip my hat to you!