Identity over the reals

Question

In the course of proving that a quadratic has two real roots, it would be very helpful to use the inequality $x^2+y^2>2xy$.

Is this inequality true for every set of two positive real numbers?

Answer 1

Your inequality is false whenever $x=y$, clearly.

If $x\neq y$ then: $$(x-y)^2>0\implies x^2-2xy +y^2>0\implies x^2+y^2>2xy$$

Answer 2

Note that $$ x^2+y^2>2xy\iff (x-y)^2=x^2+y^2-2xy>0 $$ which is true for every $x,y\in\mathbb{R}$.

Answer 3

$x^2 + y^2 >2xy \iff$

$x^2 - 2xy + y^2 > 0 \iff$

$(x-y)^2 > 0 \iff$ (assuming $x,y \in \mathbb R$)

$x-y \ne 0 \iff$

$x \ne y$.

And if $x = y$ then $x^2 + y^2 = 2x^2 = 2xy$.

So $x^2 + y^2 \ge 2xy$ for any pair of real numbers with equality holding if and only if $x = y$.