I am having trouble solving this problem. If anyone knows how, could you please help me.

Let x, y, z be positive numbers with product 1.
Prove that

(x-1+1/y)(y-1+1/z)(z-1+1/x) < 1
(less than or equal to 1)

Thank you.
Michael

OK, well first of all to make things nicer set x = a³ , y = b³ , z = c³ . We're left needing to show:

(a³ - 1 + 1/b³ )(b³ - 1 + 1/c³ )(c³ - 1 + 1/a³ ) < = 1

where abc = 1, and a,b,c are positive.

You can make the desired inequality homogeneous using the condition abc = 1, and the equality becomes:

(a³ - abc + a² c² /b)(b³ - abc + a² b² /c)(c³ - abc + b² c² /a) < = a³ b³ c³ (*)

Can you see how to get there? Basically, you replace the middle term -1 with -abc in each bracket. In the final term you replace (for instance) 1/a³ with 1/a³ * (abc)² = b² c² /a, and similarly for the other brackets.

The inequality is now in much nicer form, since each term has the same order. (In fact we don't even need the condition abc = 1 anymore. If (*) holds for positive a,b,c with abc = 1 then it must in fact hold for any positive a,b,c. Can you see why?)

Anyway, we have to prove (*). You can re-arrange it as:

(a² b - b² c + ac² )(b² c - ac² + a² b)(c² a - a² b + b² c) < = a³ b³ c³

(simply insert a factor of b/a in the first bracket, c/b in the second and a/c in the third).

Set X = a² b, Y = b² c, Z = c² a. The desired inequality is now:

(X - Y + Z)(X + Y - Z)(-X + Y + Z) < = XYZ

Note that if any of the brackets on the left are negative (or 0) then the other two must be positive, so the inequality is trivial, since the right hand side is positive and the left hand side is not.

If on the other hand all three brackets are positive then set A = X - Y + Z, B = X + Y - Z, C = -X + Y + Z and the desired result becomes:

ABC < = (A + B)/2 * (B + C)/2 * (C + A)/2

This inequality is immediate, by multiplying together the results:

sqrt(AB) < = (A + B)/2
sqrt(BC) < = (B + C)/2
sqrt(CA) < = (C + A)/2

which hold for positive A,B. (To prove them, expand out (sqrt(A) - sqrt(B))² , and its two cyclic permutations.)

Hence the inequality. Of course if this were a formal proof I should really have written this all out in reverse order.

Thank you very much for the detailed explanation to my problem.

Michael W.

Hi, I just found out this was an IMO question from 2000. There is another slightly more elegant way of going about it (which essentially boils down to the same thing).

You first prove that:

(x - 1 + 1/y)(y - 1 + 1/z) < = x² y
(y - 1 + 1/z)(z - 1 + 1/x) < = y² z
(z - 1 + 1/x)(x - 1 + 1/y) < = z² x

Can you see how to prove this? The technique relies on the fact that the difference between the right hand side and the left hand side is positive multiple of a square. Can you see how?

Once you've got those three inequalities it should be clear that if the three brackets are positive then the inequality is immediate (multiply them together and take the square root). Can you see what to do if one of the brackets isn't positive?