Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove that from sections $x,y,z$ where

$x= \sqrt[3]{(a-b)^2(a+b)}, y=\sqrt[3]{(b-c)^2(b+c)}, z=\sqrt[3]{(c-a)^2(c+a)}$ and $a,b,c>0$, $a\neq\ b \neq c$

it is possible to construct a triangle.

I started the limitation $x,y,z$ from cauchy inequality, but I am not sure if inequality I need to prove is $x+y \ge z \ge |x-y|$ and how to prove it.

share|cite|improve this question

1 Answer 1


sinc $(x^2+y^2+z^2-xy+yz+xz) >0 \implies x^3+y^3-z^3+3xyz>0 \iff x+y-z>0$

$x^3+y^3-z^3+3xyz>0 \iff -(a-b)(b-c)(a+c+2b)+3xyz>0 \iff 27(a-b)^2(b-c)^2(c-a)^2(a+b)(b+c)(a+c)> (a-b)^3(b-c)^3(a+c+2b)^3 \iff 27(c-a)^2(a+b)(b+c)(a+c) >(a-b)(b-c)(a+c+2b)^3$

if$(a-b)(b-c)<0$, then it is proved.

in case $(a-b)(b-c)>0$, note $a,c$ is symmetry, WOLG,let $c$ is min {$a,b,c$},$a=c+u,b=c+v$

$ 27(c-a)^2(a+b)(b+c)(a+c) -(a-b)(b-c)(a+c+2b)^3=(32v^2-32uv+108u^2)c^3+(48v^3-24uv^2+84u^2v+108u^3)c^2+(24v^4+9u^2v^2+75u^3v+27u^4)c+4v^5+2uv^4-3u^2v^3+11u^3v^2+13u^4v $




$\implies27(c-a)^2(a+b)(b+c)(a+c) -(a-b)(b-c)(a+c+2b)^3>0$

with same method,we have $y+z>x,x+z>y$


share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.