Prove that $a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge 0$

Question

Let $a,b$ and $c$ be the lengths of the sides of a triangle. Prove that $a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge 0$. Determine when equality occurs.

We know that if $a,b,c $ are the sides of a triangle then we can let $a=x+y$,$b=y+z$,$c=z+x$. After that I put the values of $a,b,c$ in the given expression and I found $2(xy^3+yz^3+zx^3)-2(x^2yz+xy^2z+xyz^2)$. So my target is to prove this expression is greater equal to $0$

You can use Cauchy-Schwarz Inequality for this – Mr. Xcoder May 20 '17 at 13:35 — Mr. Xcoder, May 20 '17 at 13:35

Dr. Sonnhard Graubner · Answer 1 · 2017-05-20T14:40:20.107

1

HINT: divide by $$xyz>0$$ and write $$\frac{x^2}{z}+\frac{y^2}{x}+\frac{z^2}{y}\geq x+y+z$$

edited May 20 '17 at 14:40

answered May 20 '17 at 13:25

Dr. Sonnhard Graubner

97,058

Are you sure? I think there is mistake in your computations. – Michael Rozenberg May 20 '17 at 14:26
hello Michal, do you meant me? – Dr. Sonnhard Graubner May 20 '17 at 14:30
Yes. Your dividing was wrong. – Michael Rozenberg May 20 '17 at 15:07

Prove that $a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge 0$

1 Answers1