MHB Prove √(a^2+b^2)≤ √[(x−a)^2+(y−b)^2]+√(x^2+y^2)

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The discussion centers on proving the inequality √(a² + b²) ≤ √[(x - a)² + (y - b)²] + √(x² + y²). Participants express a desire for an algebraic solution rather than a graphical representation. The need for clarity in the proof is emphasized, with some participants reiterating the request for algebraic methods. The conversation highlights the importance of understanding the geometric implications of the inequality. Ultimately, the focus remains on finding a rigorous algebraic proof for the stated inequality.
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prove:

$$\sqrt{a^2+b^2}\leq\sqrt{(x-a)^2+(y-b)^2}+\sqrt{x^2+y^2}$$
 
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solakis said:
prove:

$$\sqrt{a^2+b^2}\leq\sqrt{(x-a)^2+(y-b)^2}+\sqrt{x^2+y^2}$$

This comes from law of triangle sides

distance from origin to (a,b) < distance from (x,y) to (a,b) + distance from origin to (x,y) and equal if (x,y) is between (0,0) and (a,b) and in the line between the 2
 
kaliprasad said:
This comes from law of triangle sides

distance from origin to (a,b) < distance from (x,y) to (a,b) + distance from origin to (x,y) and equal if (x,y) is between (0,0) and (a,b) and in the line between the 2

Can you draw a picture ??

I was meaning an algebraic solution in my OP
 
solakis said:
Can you draw a picture ??

I was meaning an algebraic solution in my OP

we have $\sqrt{(x-a)^2+(y-b)^2} + \sqrt{x^2+y^2} = |(x-a) + i(y-b)| + | x + iy| = | (a-x) + (b-y) i | + | x + iy|$
$\ge |(a - x + x) + i(b-y+y)| = | a + ib| = \sqrt{a^2+b^2}$
 
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Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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