Triangle Inequality

ray giraffe

This is an olympiad style question, but it's not supposed to be very hard.

Prove that if a,b,c are the sides of a triangle then

a/(b+c-a)+b/(a+c-b)+c/(a+b-c) >= 3



If you use AMGM inequality you basically want to prove

abc >= (b+c-a)(a+c-b)(a+b-c)

which looks easier to prove. But i'm stuck.

GabbyG

You're actually done. What you got after applying the AM-GM inequality is Schur's inequality.

Schur's inequality says that for any non-negative real numbers [latex]a,b,c[/latex] and [latex]k>0[/latex]

[latex]a^{k}(a-b)(a-c)+b^{k}(b-a)(b-c)+c^{k}(c-a)(c-b) \geq 0[/latex]

The case when [latex]k=1[/latex] gives:

[latex]a^3 + b^3 + c^3 + 3abc \geq a^{2}b + a^{2}c + b^{2}a + b^{2}c + c^{2}a + c^{2}b[/latex]

which is equivalent to the inequality you got

[latex]abc \geq (b+c-a)(a+c-b)(a+b-c)[/latex]

ray giraffe

Thanks Gabby!