What is this theorem called/how to prove it?
- Thread starter miregal
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Dr.Peterson
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They can't both be true. I think you have a wrong sign in the second one. And, of course, it isn't a theorem until it's proved, so I assume you found it stated somewhere?
But try just squaring both sides, to make a start. Then you'll have to consider signs.
I have no idea what it is called.
The proof is obvious although you did not state the theorem correctly.
[MATH]\text {Let } z = \sqrt{x} + \sqrt{y} \implies z^2 = x + 2\sqrt{x} * \sqrt{y} + y \implies[/MATH]
[MATH]z^2 = x + y + 2 \sqrt{xy} = a + 2\sqrt{b} \implies z = \pm \sqrt{a + 2\sqrt{b}} \implies[/MATH]
[MATH]\sqrt{x} + \sqrt{y} = \sqrt{a + 2\sqrt{b}} \ \because \ \sqrt{x} + \sqrt{y} \ge 0.[/MATH]
The proof is obvious although you did not state the theorem correctly.
[MATH]\text {Let } z = \sqrt{x} + \sqrt{y} \implies z^2 = x + 2\sqrt{x} * \sqrt{y} + y \implies[/MATH]
[MATH]z^2 = x + y + 2 \sqrt{xy} = a + 2\sqrt{b} \implies z = \pm \sqrt{a + 2\sqrt{b}} \implies[/MATH]
[MATH]\sqrt{x} + \sqrt{y} = \sqrt{a + 2\sqrt{b}} \ \because \ \sqrt{x} + \sqrt{y} \ge 0.[/MATH]