IIT JEE , AIEEE Forum - Mathematics , Algebra - this question is given to me by my friend ,its a bit crooked but can see no way of solving

vineetnegi

x+y+z=2

2^(x^2+x) + 2^(y^2+y) +2^(z^2+z) =6*(2^1/9)

solve in {R}

akhil_o

sorry...edited...didint see frst condition

getting no real solution

vineetnegi

i am sorry for mistyping the question
....review it

akhil_o

solution is (2/3,2/3,2/3)

anchitsaini

R.H.S=
3*2*2^1/9
=3*2^10/9
as on the l.h.s are 3 terms
we can proceed taking
x^2+x=y^2+y=z^2+z=10/9
the above eqn gives soln
2/3 and -5/3

and for the given condition
x+y+z=2
the ans is
2/3,2/3,2/3

akhil_o

yeah i was thinking along same lines...

RHS has 3 term so LHS has to be divisible by 3
so all three terms on lhs must leave same rem when divided by 3
and as x+y+z=2
least soln is taken...equality case
solve the uadratic and get 2/3

hsbhatt

I was getting this 2^1/9 all morning and wanted to ask you but then thought better of it. Now that you have changed the exponent, here is the solution:

We have by Cauchy-Schwarz inequality(or Chebyshev), 3(x²+y²+z²)

(x+y+z)²

Hence x²+y²+z²

4/3

Now let a = 2^{x^2+x}; b = 2^{y^2+y}; c= 2^{z^2+z}

From AM-GM inequality, (a+b+c)/3

³

abc

abc = 2^{x^2+x} . 2^{y^2+y} . 2^{z^2+z} = 2^{x2+y2+z2 +x+y+z}

2^4/3+2

2^10/3

Hence (a+b+c)/3

³

abc

2^10/9

Hence a+b+c

3.2^10/9 = 6.2^1/9

But, we are given a+b+c = 6.2^1/9

So, all we have to do is to find under what condition the equality holds.

For the Cauchy-Schwarz inequality, the condition is x=y=z, which luckily satisfies the condition for the AM-GM inequality with a,b and c

Hence x = y = z = 2/3 is the only solution set for the equation.

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