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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.

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1 Answer 1

$x^3+y^3-z^3+3xyz=(x+y-z)(x^2+y^2+z^2-xy+yz+xz)$

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 $

$32v^2-32uv+108u^2>0$

$48v^3-24uv^2+84u^2v+108u^3>0$

$4v^5+2uv^4-3u^2v^3+11u^3v^2+13u^4v>0$

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

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

QED.

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