Prove $a^2 + b^2 \geq 2ab$ using Triangle Inequality

Question

Let $a$ and $b$ be two real, positive numbers. Is it possible to prove $$a^2+b^2 \ge 2ab$$ using the Triangle Inequality?

This was suggested to me as a proof method but I have been unsuccessful so far.

Answer 1

By the triangle inequality $$|a-b|+|b-a|\geq|a-b+b-a|=0.$$ Thus, $$|a-b|\geq0$$ or $$(a-b)^2\geq0$$ or $$a^2+b^2\geq2ab.$$