1Deepak bhaiya is correct. but it was for class 9, 10 students. it can also be solved using cauchy schwarz(try it!)
prove that \sqrt{(x_1-x_2)^2+ (y_1-y_2)^2}\le \sqrt{x^2_1+y^2_1}+ \sqrt{x^2_2+y^2_2}
for all real numbers x_1,y_1,x_2,y_2
HINT:proceed by forming a triangle with suitable coordinates
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10 Answers
21assume triangle with cordinates as origin(0,0) (x1,y1) (x2,y2)
we will get this inequality by the fact that sum of lengths of two sides of triangle is always greater than length of third side
121oops.........i din see that.. sorry....and dont call me bhaiya :P.....but seriously how many here on tiit r in class 9-10.......i guess no one :P ....
1@theprophet sir thanks for the information.
is anybody trying to solve this using cauchy??
23wat is cauchy and minkowski inequality?[7]? i hav never come across dem b4 ever [2]...pls post it if it is helpful....[1]
1can any body explain minkowski inequality in a simpler way......i saw it in wiki but everything went above my head.....
1and can anybody post all such iequalities which r useful for jee
1(x_1-x_2)^2+(y_1-y_2)^2=x^2_1+x^2_2+y_1^2+y_2^2+2(x_1.x_2+y_1.y_2) ----------------(1)
from cauchy schwarz, we know
(x_1.x_2+y_1.y_2)^2 \le (x_1^2+x_2^2)(y_1^2+y_2^2)
or (x_1.x_2+y_1.y_2)\le \sqrt{(x_1^2+x_2^2)(y_1^2+y_2^2)}
from (1) we get
(x_1-x_2)^2+ (y_1-y_2)^2 \le \left( \sqrt{x_1^2+x_2^2}+\sqrt{y_1^2+y_2^2}\right)^2
taking the square root on both sides, we get the desired inequality
1squaring we get,
-(x1x2+y1y2)≤2[√x12+y12 + √x22+y22
again squaring,
(x1y2-x2y1)2≥0
which is always true
therefore the given equation is also correct.