AlgebraDifficulty 7.9National olympiad, round 2Find the answer
Find all integer n such that the following property holds: for any positive real numbers a,b,c,x,y,z, with max(a,b,c,x,y,z)=a , a+b+c=x+y+z and abc=xyz, the inequality an+bn+cn≥xn+yn+zn 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, a+b+c=x+y+z, and abc=xyz. We need to find all integer n such that the inequality an+bn+cn≥xn+yn+zn holds for any positive real numbers a,b,c,x,y,z.
We claim that the answer is all n≥0.
### Proof:
1. **Case n<0**: For n<0, consider the counterexample: (a,b,c)=(2,3,4),(x,y,z)=3.5,25.5+5.52−74⋅48,25.5−5.52−74⋅48. This shows that an+bn+cn≥xn+yn+zn is not necessarily true for n<0.
2. **Case n≥0**: We will show that for all n≥0, the inequality holds.
Define p=abc and s=a+b+c. Let S be the set of possible values attained by max(a,b,c) as a,b,c∈R+ vary while satisfying abc=p and a+b+c=s.
Lemma 1: The set S is a closed interval. - This follows from the fact that a uniquely determines b and c via the quadratic equation derived from a+b+c=s and abc=p.
Lemma 2: As r varies in S, define f(r) as the value of ab+bc+ca when a=r. Then f(r) is nonincreasing on S. - This is shown by expressing f(r) and proving that its derivative is nonpositive.
Define di=ai+bi+ci and wi=xi+yi+zi for all i∈Z. Set ti=di−wi.
Using the recurrence relations and the properties of f(r), we derive that: tn≥(x+y+z)tn−1−(xy+yz+zx)tn−2+xyztn−3.
By induction, starting from t0=t1=0 and t2≥0, we show that tn≥0 for all n≥0.
Therefore, the inequality an+bn+cn≥xn+yn+zn holds for all n≥0.
The answer is: n≥0.
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