Inequality

Let a, b, and c be positive real numbers.

Prove that $a^2 + b^2 + c^2 >= ab + ac + bc$.

$\large{AM\ge GM}$

$\begin{array}{|rcll|} \hline \mathbf{\dfrac{a+b}{2}} &\ge& \mathbf{\sqrt{ab}} \\ a+b &\ge& 2\sqrt{ab} \quad &| \quad \text{square both sides} \\ (a+b)^2 &\ge& 4ab \\ a^2+2ab+b^2 &\ge& 4ab \quad &| \quad -2ab \\ a^2+b^2 &\ge& 4ab-2ab \\ \mathbf{a^2+b^2} &\ge& \mathbf{2ab} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \mathbf{\dfrac{b+c}{2}} &\ge& \mathbf{\sqrt{bc}} \\ b+c &\ge& 2\sqrt{bc} \quad &| \quad \text{square both sides} \\ (b+c)^2 &\ge& 4bc \\ b^2+2bc+c^2 &\ge& 4bc \quad &| \quad -2bc \\ b^2+c^2 &\ge& 4bc-2bc \\ \mathbf{b^2+c^2} &\ge& \mathbf{2bc} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \mathbf{\dfrac{a+c}{2}} &\ge& \mathbf{\sqrt{ac}} \\ a+c &\ge& 2\sqrt{ac} \quad &| \quad \text{square both sides} \\ (a+c)^2 &\ge& 4ac \\ a^2+2ac+c^2 &\ge& 4ac \quad &| \quad -2ac \\ a^2+c^2 &\ge& 4ac-2ac \\ \mathbf{a^2+c^2} &\ge& \mathbf{2ac} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{a^2+b^2} &\ge& \mathbf{2ab} \\ \mathbf{b^2+c^2} &\ge& \mathbf{2bc} \\ \mathbf{a^2+c^2} &\ge& \mathbf{2ac} \\ \hline (a^2+b^2)+(b^2+c^2) + (a^2+c^2) &\ge& 2ab + 2bc +2ac \\ 2a^2+2b^2+2c^2 &\ge& 2(ab +bc +ac)\\ 2(a^2+b^2+c^2) &\ge& 2(ab +bc +ac) \quad &| \quad : 2\\ \mathbf{a^2+b^2+c^2} &\ge& \mathbf{ab +bc +ac} \\ \hline \end{array}$

Under what conditions does equality occur? 

$\begin{array}{|rcll|} \hline 2(a^2+b^2+c^2) &=& 2(ab +bc +ac) \\ 2(a^2+b^2+c^2)- 2(ab +bc +ac) &=& 0 \\ \hline \end{array}\\ 2(a^2+b^2+c^2)- 2(ab +bc +ac) =(a-b)^2+(b-c)^2+(c-a)^2 \\ \begin{array}{|lcll|} \hline (\underbrace{a-b}_{=0})^2+(\underbrace{b-c}_{=0})^2+(\underbrace{c-a}_{=0})^2&=& 0 \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline a-b&=& 0 \\ a&=& b \\\\ b-c&=& 0 \\ b&=& c \\\\ c-a &=& 0 \\ c&=& a \\ \hline \end{array}$

equality occur when $a=b=c$