Inequality show that question involving two equations

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The discussion centers on proving the inequality |x| + |y| ≥ √(x^2 + y^2) for all real values of x and y. The user contemplates squaring both sides to simplify the proof, leading to the conclusion that 2|xy| ≥ 0, which is always true. The equality holds when either x or y is zero, as indicated by |xy| = 0. The conversation highlights the challenge of manipulating two variables simultaneously in the context of the inequality. Ultimately, the proof confirms the validity of the inequality and identifies the conditions for equality.
blopblop
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Hi, so here is my question that I am totally stumped on.

for all real values of x and y, show that |x|+|y|≥ √(x^2+y^2 )

and find the real values of x and y in which equality holds.


I sort of thought I could do the second part, but it confuses me with two pronumerals and how to get rid of one so that I can find the other.
 
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Both sides are positive so the inequality must hold if and only if the square of both sides satisfies the same inequality. That might simplify it a bit.
 
|x| + |y| ≥ √(x*x +y*y) / ( )2 squaring

x*x + y*y + 2*|x*y| ≥ x*x + y*y /-x*x,-y*y

2*|x*y| ≥ 0 /:2

|x*y| ≥ 0 It is true.


|x*y| = 0 ↔ x = 0 or y = 0

kamke
 
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