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Algebra Difficulty 7.9 National olympiad, round 2 Find the answer

Find all integer nn such that the following property holds: for any positive real numbers a,b,c,x,y,za,b,c,x,y,z, with max(a,b,c,x,y,z)=amax(a,b,c,x,y,z)=a , a+b+c=x+y+za+b+c=x+y+z and abc=xyzabc=xyz, the inequality an+bn+cnxn+yn+zna^n+b^n+c^n \ge x^n+y^n+z^n holds.

A number or a short expression. Spacing and $ signs are ignored.

Solution

We are given the conditions max(a,b,c,x,y,z)=a \max(a, b, c, x, y, z) = a , a+b+c=x+y+z a + b + c = x + y + z , and abc=xyz abc = xyz . We need to find all integer n n such that the inequality
an+bn+cnxn+yn+zn a^n + b^n + c^n \ge x^n + y^n + z^n
holds for any positive real numbers a,b,c,x,y,z a, b, c, x, y, z .

We claim that the answer is all n0 n \ge 0 .

### Proof:

1. **Case n<0 n < 0 **:
For n<0 n < 0 , consider the counterexample:
(a,b,c)=(2,3,4),(x,y,z)=(3.5,5.5+5.5244872,5.55.5244872). (a, b, c) = (2, 3, 4), \quad (x, y, z) = \left(3.5, \frac{5.5 + \sqrt{5.5^2 - \frac{4 \cdot 48}{7}}}{2}, \frac{5.5 - \sqrt{5.5^2 - \frac{4 \cdot 48}{7}}}{2}\right).
This shows that an+bn+cnxn+yn+zn a^n + b^n + c^n \ge x^n + y^n + z^n is not necessarily true for n<0 n < 0 .

2. **Case n0 n \ge 0 **:
We will show that for all n0 n \ge 0 , the inequality holds.

Define p=abc p = abc and s=a+b+c s = a + b + c . Let S S be the set of possible values attained by max(a,b,c) \max(a, b, c) as a,b,cR+ a, b, c \in \mathbb{R}^+ vary while satisfying abc=p abc = p and a+b+c=s a + b + c = s .

Lemma 1: The set S S is a closed interval.
- This follows from the fact that a a uniquely determines b b and c c via the quadratic equation derived from a+b+c=s a + b + c = s and abc=p abc = p .

Lemma 2: As r r varies in S S , define f(r) f(r) as the value of ab+bc+ca ab + bc + ca when a=r a = r . Then f(r) f(r) is nonincreasing on S S .
- This is shown by expressing f(r) f(r) and proving that its derivative is nonpositive.

Define di=ai+bi+ci d_i = a^i + b^i + c^i and wi=xi+yi+zi w_i = x^i + y^i + z^i for all iZ i \in \mathbb{Z} . Set ti=diwi t_i = d_i - w_i .

Using the recurrence relations and the properties of f(r) f(r) , we derive that:
tn(x+y+z)tn1(xy+yz+zx)tn2+xyztn3. t_n \ge (x + y + z) t_{n-1} - (xy + yz + zx) t_{n-2} + xyz t_{n-3}.

By induction, starting from t0=t1=0 t_0 = t_1 = 0 and t20 t_2 \ge 0 , we show that tn0 t_n \ge 0 for all n0 n \ge 0 .

Therefore, the inequality an+bn+cnxn+yn+zn a^n + b^n + c^n \ge x^n + y^n + z^n holds for all n0 n \ge 0 .

The answer is: n0\boxed{n \ge 0}.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.