Maths Proof Help?

Question

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!

? · Accepted 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)

.

? · Answer

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.

DWRead · Answer

Do you mean x + y ≥ √(x^2 + y^2) ?

Maths Proof Help?

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