TITLE: Prove that $ { a }^{ 2 }+2ab+{ b }^{ 2}\ge 0$ without using $(a+b) ^{ 2 }$
QUESTION [5 upvotes]: Prove that $${ a }^{ 2 }+2ab+{ b }^{ 2 }\ge 0,\quad\text{for all }a,b\in \mathbb R $$
  without using $(a+b)^{2}$.

My teacher challenged me to solve this question from any where. He said you can't solve it. I hope you can help me to solve it.

REPLY [4 votes]: Case 1. If both $a,b$ are positive , then each term $a^2, 2ab, b^2$, all are positive. Hence $a^2 + 2ab +b^2>0$. 
Case 2. If both $a,b$ are negative, then also each term $a^2, 2ab, b^2$, all are positive. Hence $a^2 + 2ab +b^2>0$. 
Case 3. If one of $a,b$ is positive and other is negative, then in this case by changing one of its sign sufficient to show $a^2 -2ab+b^2 > 0$, where $a,b$ both are positive. So, without loss of generality assume $a>b$. 
Here is a pictorial proof of that fact.