Prove that if {a;b} $\in \mathbb R^+$ then $a^2+b^2>ab$

Question

I have tried factoring it already, but it doesn't seem to evolve much:

First I multiply each side by $2$:

$ 2(a^2+b^2)>2ab$

Then I substitute using the relation $(a+b)^2=a^2+2ab+b^2$ and it becomes:

$2(a^2+b^2)>(a+b)^2 - (a^2+b^2)$

and then:

$3(a^2+b^2)>(a+b)^2$

And that's pretty much it, I'm stuck.

Answer 1

We have that

$$0\le (a-b)^2=a^2+b^2-2ab.$$ Thus we have

$$2ab\le a^2+b^2.$$

Now, if $ab$ is positive then

$$ab< 2ab\le a^2+b^2.$$ And if $ab$ is negative then

$$ab< 0 <a^2+b^2.$$

Answer 2

For $ab\ge0$:

$$(a-b)^2 \ge 0 \Rightarrow a^2+b^2 \ge 2ab >ab$$

If $ab<0$ then we can easily write

$$a^2+b^2>ab$$

because the left side is positive and the right side is negative.

Answer 3

ab < max(a,b) * max(a,b)

Let's say that a>b ab < aa = a^2 < a^2 + b^2 since b is not null (R) if b

Answer 4

$(a-b)^2\ge 0$

$\implies a^2+b^2\ge 2ab$

$\implies a^2+b^2\gt ab$

if $ab$ is negative then it follows as LHS is always positive

Answer 5

$$ a^2+b^2-ab=\frac{(2a-b)^2+3b^2}{4}. $$

Answer 6

$$ (a-b)^2 \geq 0 \\ (a-b)^2 = a^2 - 2ab + b^2 \\ a^2 + b^2 \geq 2ab \geq ab $$