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

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SUMMARY

The inequality $$\sqrt{a^2+b^2}\leq\sqrt{(x-a)^2+(y-b)^2}+\sqrt{x^2+y^2}$$ is established as a geometric representation of the triangle inequality in Euclidean space. The proof involves demonstrating that the distance from the origin to the point (a, b) is less than or equal to the sum of the distances from the origin to (x, y) and from (x, y) to (a, b). This conclusion is supported by algebraic manipulation and geometric interpretation, confirming the validity of the 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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