good job anchit. This was the soln I was looking for which gives the answers neatly.

 

 

Many thanks for using solution by inspection.

 

There are some more solutions, pls find them too.

Find xR such that sin³x+cos³x = 2 sinx cosx.

 

And if you have the time xC also!

Preamble: I am posting this bcos anchit insisted. Strictly no rates not only bcos i filched the soln, but also bcos i am not capable of solving such a problem. Rates would embarass me.

 

Solution:

It is given that ab+bc+ca3k²-1

 

WLOG a>b>c

 

Hence a-b>1, b-c>1 and a-c>2

 

Now a²+b²+c²-ab-bc-ca = 0.5*[(a-b)² + (b-c)² + (c-a)²]  0.5*(1+1+4] = 3.

 

Hence a³+b³+c³ - 3abc = 0.5*(a+b+c) *[(a-b)² + (b-c)² + (c-a)²]  3(a+b+c)

 

Now (a+b+c)² = a²+b²+c²+2(ab+bc+ca)

 

= a²+b²+c²-(ab+bc+ca)+3(ab+bc+ca)3+3(3k²-1) = 9k²

 

Hence a+b+c3k

 

Hence a³+b³+c³ - 3abc9k

 

a³+b³+c³-3abc = (a+b+c) (a+b+c²) (a+b²+c)

 

and

 

x³+y³+z³-3xyz = (x+y+z) (x+y+z²) (x+y²+z)

 

Hence

Hint:

a³+b³+c³-3abc = (a+b+c) (a+b+c²) (a+b²+c)

I dont think we have the patience to decode your answer. The administrators have provided such a helpful editor. use it man.

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  ³abc

 

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  ³abc  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.

 

There is one more approach, and this line of reasoning could also prove useful:

 

Again rewrite as 2004^x - 2003^x = 2002^x - 2001^x 

 

Now, consider f(t) = t^x.

 

By Intermediate Value Theorem, there exists p[2004,2003] such that

 

xp^x-1 =  2004^x - 2003^x /(2004-2003) = 2004^x - 2003^x

 

Similarly there exists q[2004,2003] such that

 

xq^x-1 =  2002^x - 2001^x /(2002-2001) = 2002^x - 2001^x

 

Hence xp^x-1 = xq^x-1

 

or  x(p^x-1 -q^x-1) = 0

 

Since p and q are distinct, x=0 or x=1 are the only possibilities.

Method I:

 

Write the equation as 2004^x - 2003^x = 2002^x - 2001^x 

 

Now, consider the function f(x) = p^x - (p-1)^x for p>1.

 

f'(p) = x(p^x-1 - (p-1)^x-1)

 

You can see that f'(x) = 0 for x=0 or x=1.

 

for any other value of x, f'(p) is a strictly monotonic function.

 

which means 2004^x - 2003^x = 2002^x - 2001^x only if x=0 or x =1.

 

 

 

 

Assuming sin and cos are +ve , it is a straightforward application of Cauchy-Schwarz inequality:

 

(a²+b²) (c²+d²)  (ac+bd)²

 

If a = 1, b = 1/sin^n/2, c = 1, d = 1/cos^n/2

 

then (1+1/sinⁿ) (1+1/cosⁿ)  (1+1/sin^n/2cos^n/2)² = (1+2^n/2/sin^n/22)²  (1+2^n/2)²

 

Equality occurs when  = n + /4

 

Call conjugate of z as z'

 

Rewriting the eqn as zz'w - ww'z = z-w gives z(1-z'w) = w(1-w'z).

 

One solution is z'w = 1 = zw'.

 

If z'w 1,

 

1-z'w and 1-w'z are conjugates and so |1-z'w | = |1-w'z|  0

 

Hence |z| = |w|.......Eqn2

 

Hence |z|²w -|w|²z = z-w implies |z|² (w-z) = z-w

 

Now, necessarily  z=w as zw would mean |z|² = -1 which is impossible.

 

Hence if |z|²w -|w|²z = z-w either z'w = zw' =1 or z=w.

 

Proving that if z'w = zw' =1 or z=w then |z|²w -|w|²z = z-w is easy.

Good. Any other approaches. You can take a hint from the category under which it is posted.

 

I said any other approach, because the above one is not completely legitimate. I dont think all of us are comfortable with binomial theorem applied to real numbers. There is something more to be done

Find x such that

 

2001^x+2004^x = 2002^x+2003^x