Seems no one is interested in doing good level IIT problems but all are up for out of course olympiad probs.

Does it have DPPs? and where are you from?

Adding 3 to both sides we get:

(a+b +c) (1/b+c  +  1/c+a  + 1/a+b) >= 9/2

or

2(a+b +c) (1/b+c  +  1/c+a  + 1/a+b) >= 9

or

((b+c) + (c+a) + (a+b)) (1/b+c  +  1/c+a  + 1/a+b) > =9

which is true by AM >=HM

Hence proved , no use of a+b+c =1 and no hardcore inequalities this is the best way to prove the Nesbitt's inequality.

Err, I missed the x but I edited my post after 30 seconds or so. Seems you didnt see the correction.

Also the answer is numeric only.

A question done in my class

x^2 - 2(a^2 + b^2 + c^2)x +3L(ab +bc +ca) =0

where a,b,c represent sides of a triangle, then find the condition on L for the equation to have real roots.

^ No , plz dont hit flukes. Infact the condition we need comes out to be independent of @.

Its any angle. This is a very short question.

A very nice question.

a,b are complex nos.Given , a^2 + b^2 + 2abcos@ =0

Then the triangle with vertices a,b,0 is

a)Right

b)Isoceles

c)Scalene

d)Equilateral

Its perfectly all right.. Infact the field due to the plane containing the circular loop adds up to 0. Look closely the current is divided there.

No, potential energy is defined as -integral F.dx , where F is conservative .

Its for calculation purposes only.

Why create 3 threads for one question huh?

If a>0 , there is definitely a root.

Did you mean a<0 in the question?

Chill

http://olympiads.hbcse.tifr.res.in/enrollment

www.artofproblemsolving.com

Sir can you explain c) a little more in detail?