IIT JEE , AIEEE Forum - Mathematics , Trignometry

anandghegde

If a, b, c are the sides of triangle ABC, such that

$(1 + \frac{b-c}{a})^a (1 + \frac{c-a}{b})^b (1 + \frac{a-b}{c})^c \ge 1$

then ABC must be
a] right triangle
b] isoceles
c] obtuse angled
d] equilateral

Radon222

anandghegde

wrong!

digs2digi

is the answer c??????/
cant be eqilateral or isosceles as that wud give 0>=1.
and i think it cannot be right angled also

digs2digi

or maybe it can be right angled.i havent checked

digs2digi

PLZ IGNORE MY LAST 2 POSTS. I NOTED THE QUES INCORRECTLY.SRY

hash_include

answer
D equilateral.. the value can ONLY be 1 and not > 1

anandghegde

please post your soln dude....

hash_include

well, i didn't do any rigorous method to solve it..
since the question is given 'MUST be', then a good way is to substitute values, which is what i did..
and since only an equilateral triangle is possible, thats the answer!

sboosy

given is

((a+b-c)/a) ^a ((b+c-a)/b)^b ((c+a-b)/c)^c ..... = H

consider

(a+b-c)/a + (a+b-c)/a + (a+b-c)/a .....a times

similarly

(b+c-a)/b + (b+c-a)/b + .....b times

similarly

(c+a-b)/c + (c+a-b)/c + ...c times

taking arithmetic mean of them

[a (a+b-c)/a + b(b+c-a)/b + c (c+a-b)/c ]/(a+b+c) >= H ^1/(a+b+c)

but LHS is 1

so

H< = 1

but given is H> = 1

so H = 1

now comes the point

u may still argue that each bracket need not b 1

but look at this

b+c>a(sum of 2 sides greater than third)

so a-b<c

(a-b)/c <1

adding 1

1+(a-b)/c < 2

now if one of the brackets is not =1

then other shud b greater than 2 for 1 to come

since this is not possible each bracket = 1

so b-c = 0

c-a = 0

so a = b = c

hence equilateral

sandeepramesh

actually the answer shd be d i believe bcos the inequality goes like for a triangle,

$ (1+ \frac{b-c}{a})^a \cdot (1+ \frac{c-a}{b})^b \cdot (1+ \frac{a-b}{c})^c leq 1$

proof:

AM $geq$ GM

$implies$ $((1+ \frac{b-c}{a}) + (1+ \frac{b-c}{a}) +.........+ a times) + $((1+ \frac{c-a}{b}) + (1+ \frac{c-a}{b}) +.........+ b times) ...............................

----------------------------------------------------------------------------

(a+b+c)

$geq$ the GM But we get AM = 1

Hence its equilateral and answer is D

hsbhatt

Sandeep's approach is nice.

Another way:

In a triangle with sides a,b, and c, the following inequality holds

abc

(a+b-c)(b+c-a)(c+a-b)

This is because a = 1/2 [(a+b-c) + (c+a-b)]

(a+b-c)(a+c-b) etc.

Hence, (1+b-c/a) (1+c-a/b) (1+a-b/c)

1

with equality when a =b =c

Now WLOG a>b>c

Hence (1+b-c/a)^a (1+c-a/b)^b (1+a-b/c)^c

(1+b-c/a)^a (1+c-a/b)^a (1+a-b/c)^a = [(1+b-c/a) (1+c-a/b) (1+a-b/c)]^a

1.

Again equality holds when a=b=c.

Since here, (1+b-c/a)^a (1+c-a/b)^b (1+a-b/c)^c

1, we must have a=b=c

sandeepramesh

hsbhatt im madness :)

hsbhatt

no wonder! I was thinking, we have a really good newbie

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