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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