Prove inequality 2ab <= a^2 + b^2 from 0 <= (a - b)^2

[jetoso]

Show that: 2ab <= a^2 + b^2

It follows that 0 <= (a - b)^2 is going to be always positive, then inequelity holds. But I think I need to prove that 2 times a times b has to be less than equal to the sum of the squares of a and b.
Any suggestions?

 

[amcavoy]

Show that: 2ab <= a^2 + b^2

It follows that 0 <= (a - b)^2 is going to be always positive, then inequelity holds. But I think I need to prove that 2 times a times b has to be less than equal to the sum of the squares of a and b.
Any suggestions?

I believe that's all you need to do (expand (a-b)²).

$$\left(a-b\right)^2\geq 0\implies a^2+b^2\geq 2ab$$

...so I think your initial argument is proof enough.

 

[TD]

Well, you practically solved it yourself, no?

$$\left( {a - b} \right)^2 \ge 0 \Leftrightarrow a^2 - 2ab + b^2 \ge 0 \Leftrightarrow a^2 + b^2 \ge 2ab$$