Inequality show that question involving two equations

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SUMMARY

The inequality |x| + |y| ≥ √(x² + y²) holds for all real values of x and y, as demonstrated through algebraic manipulation. By squaring both sides, the expression simplifies to 2|xy| ≥ 0, which is always true. Equality occurs when either x = 0 or y = 0, indicating that at least one of the variables must be zero for the equality to hold.

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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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