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

[solakis1]

prove:

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

 

[kaliprasad]

prove:

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

Spoiler

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]

Spoiler

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]

Can you draw a picture ??

I was meaning an algebraic solution in my OP

Spoiler

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