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

solakis1 · May 5, 2017

prove:

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

kaliprasad · May 6, 2017

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

solakis1 · May 7, 2017

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

kaliprasad · May 7, 2017

solakis said:

Can you draw a picture ??

I was meaning an algebraic solution in my OP

we have $\sqrt{(x-a)^2+(y-b)^2} + \sqrt{x^2+y^2} = |(x-a) + i(y-b)| + | x + iy| = | (a-x) + (b-y) i | + | x + iy|$
$\ge |(a - x + x) + i(b-y+y)| = | a + ib| = \sqrt{a^2+b^2}$

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

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