IIT JEE , AIEEE Forum - Mathematics , Algebra

vineetnegi

if a,b,c are sides of triangle

proove

a^2(p-q)(p-r)+b^2(q-r)(q-p)+c^2(r-p)(r-q)

cannot be negative;p,q,r being any real quantities

(higher algebra :H.S halls)

vineetnegi

aray! koi to batao

hsbhatt

IF a

b

c and p

q

r

a²(p-q)(p-r)+b²(q-r)(q-p)+c²(p-r)(p-q)

b²(p-q)(p-r) + b²(q-r)(q-p) + c²(p-r)(p-q)

= b²(p-q)²+c²(p-r)(p-q)

0

But, I am not sure we can assume WLOG that a

b

c and p

q

r. Got to see.

konichiwa2x

Let

.

,and

such that

and inequality to prove become

, similarly:

Now the requirement that x+y+z = 0 guarantees that for (x,y,z) one has
(0,0,0) (0, +, - ) (+,+,-) or (+, - , - ) as positive/negative quantities for the triple!
Thus atleast one of xy,xz, or yz has a product which is positive!

Finally, the triangle inequality ensures the following: a+b>c, a+c>b, and b+c>a
Combining all the above assures us that at least one of the above representations for Q is the sum of two positive quantities.

hsbhatt

Actually I should kick myself for not having finished this off earlier:

a²(p-q)(p-r)+b²(q-r)(q-p)+c²(r-p)(r-q)

m² [(p-q)(p-r)+(q-r)(q-p)+(r-p)(r-q)] where m = min(a,b,c)

Now (p-q)(p-r)+(q-r)(q-p) = (p-q)(p-r+r-q) = (p-q)². Similarly

(q-r)(q-p) + (r-p)(r-q) = (q-r)² and

(r-p)(r-q)+(p-q)(p-r) = (p-r)²

Adding these three, we get [(p-q)(p-r)+(q-r)(q-p)+(r-p)(r-q)] = 1/2 [(p-q)²+(q-r)²+(r-p)²]

Hence a²(p-q)(p-r)+b²(q-r)(q-p)+c²(r-p)(r-q)

m² 1/2 [(p-q)²+(q-r)²+(r-p)²]

0

Only i have not used that they are sides of a triangle.

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