Answer:
Proof in explanation.
Step-by-step explanation:
I'm going to assume you mean "greater than or equal to".
We know [tex](x-y)^2 \ge 0[/tex].
We know this because any real number squared will result in a positive or zero result.
Let's expand the left hand side:
[tex](x-y)(x-y) \ge 0[/tex]
Distribute:
[tex]x(x-y)-y(x-y) \ge 0[/tex]
[tex]x^2-xy-yx+y^2 \ge 0[/tex]
Combine like terms:
[tex]x^2-2xy+y^2 \ge 0[/tex]
Commutative property of addition:
[tex]x^2+y^2-2xy \ge 0[/tex]
Add [tex]2xy[/tex] on both sides:
[tex]x^2+y^2 \ge 2xy[/tex]
Now we wanted to show [tex]x^2+y^2 \ge xy[/tex]. Our right hand side almost appeared that way except the factor of 2 present there.
Let's go back to our inequality that we got:
[tex]x^2+y^2 \ge 2xy[/tex]
Divide both sides by 2:
[tex]\frac{x^2+y^2}{2} \ge xy[/tex]
Note: Half of a positive number is less than that positive number.
[tex]x^2+y^2 \ge \frac{x^2+y^2}{2} \ge xy[/tex]
This inequality says we would like it to say in the end:
[tex]x^2+y^2 \ge xy[/tex]
