solakis1
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prove:
$$\sqrt{a^2+b^2}\leq\sqrt{(x-a)^2+(y-b)^2}+\sqrt{x^2+y^2}$$
$$\sqrt{a^2+b^2}\leq\sqrt{(x-a)^2+(y-b)^2}+\sqrt{x^2+y^2}$$
solakis said:prove:
$$\sqrt{a^2+b^2}\leq\sqrt{(x-a)^2+(y-b)^2}+\sqrt{x^2+y^2}$$
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
solakis said:Can you draw a picture ??
I was meaning an algebraic solution in my OP