# Abstract Inequalities

• Oct 21st 2012, 11:38 AM
GrigOrig99
Abstract Inequalities
Q. Having some trouble with this one. Can anyone help me out?

Many thanks.

Q. Use the result, $a^2+b^2\geq2ab$, to prove that $a^2+b^2+c^2\geq ab+bc+ca$ for all real values of a, b & c.

Attempt: $\frac{a^2+b^2}{2}\geq ab$
if $\frac{a^2+b^2}{2}-ab\geq0$
if $\frac{(a+b)(a-b)}{2}-ab\geq0$
if $\frac{a^2-ab+ab+b^2}{2}-ab\geq0$
if $\frac{a^2-ab+ab+b^2-2ab}{2}\geq0$
if $\frac{a^2-2ab+b^2}{2}$
if $\frac{(a-b)^2}{2}\geq0$

Thus: $a^2+b^2+c^2\geq ab+bc+ca$
if $\frac{2a^2+2b^2+2c^2}{2}-ab-bc-ca\geq0$
if $\frac{2a^2+2b^2+2c^2-2ab-2bc-2ca}{2}\geq0$
if $a^2-b^2-c^2-ab-bc-ca\geq0$
if $(a+b+c)^2\geq0$
• Oct 21st 2012, 11:52 AM
Plato
Re: Abstract Inequalities
Quote:

Originally Posted by GrigOrig99
Q.
Q. Use the result, $a^2+b^2\geq2ab$, to prove that $a^2+b^2+c^2\geq ab+bc+ca$ for all real values of a, b & c.

\begin{align*} a^2+b^2 &\ge 2ab \\ a^2+c^2 &\ge 2ac \\ b^2+c^2&\ge 2bc \\ \text{Add all three }2a^2+2b^2+2c^2 &\ge 2ab+2ac+2bc \end{align*}
• Oct 21st 2012, 12:24 PM
GrigOrig99
Re: Abstract Inequalities
Great, thank you very much.