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Maths Proof Help?
How do I prove that x + y ≥ √x^2 + y^2 if x and y are both positive.
I know I can prove it with certain examples, but I need to carry out a proof that proves it for all cases.
Any help is greatly appreciated!
x + y ≥ √(x^2 + y^2)
3 Answers
- hiiLv 63 years agoFavorite Answer
(1)
x > 0, y > 0, x + y > √(x² + y²)
(x + y)² - [√(x² + y²)]²
= (x² + 2xy + y²) - (x² + y²)
= 2xy > 0
then x + y > √(x² + y²)
(2)
x ≥ 0, y ≥ 0, x + y ≥ √(x² + y²)
(x + y)² - [√(x² + y²)]²
= (x² + 2xy + y²) - (x² + y²)
= 2xy ≥ 0
then x + y ≥ √(x² + y²)
.
- az_lenderLv 73 years ago
Parentheses are REQUIRED, to show that the RHS means sqrt(x^2+y^2) rather than sqrt(x) + y^2.
Here's a proof: square both sides:
x^2 + 2xy + y^2 >= x^2 + y^2 ?
Yes, of course; if x and y are positive, then 2xy is positive, and it's for sure that x^2 + 2xy + y^2 > x^2 + y^2.
