1. ## Abstract Inequalities

Q. Having some trouble with this one. Can anyone help me out?

Many thanks.

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

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

Thus: $\displaystyle a^2+b^2+c^2\geq ab+bc+ca$
if $\displaystyle \frac{2a^2+2b^2+2c^2}{2}-ab-bc-ca\geq0$
if $\displaystyle \frac{2a^2+2b^2+2c^2-2ab-2bc-2ca}{2}\geq0$
if $\displaystyle a^2-b^2-c^2-ab-bc-ca\geq0$
if $\displaystyle (a+b+c)^2\geq0$

2. ## Re: Abstract Inequalities

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

3. ## Re: Abstract Inequalities

Great, thank you very much.