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Silk Road Mathematics Competition competition problems

Algebra Difficulty 6.3 National Olympiad Prove it Silk Road Mathematics Competition

Let nn be an integer with n>2n > 2 and a1,a2,…,an∈R+a_1, a_2, \dots, a_n \in \mathbb{R}^+ positive real numbers. Given any positive integers t,k,pt, k, p with 1<t<n1 < t < n, set m=k+pm = k+p. Prove the following inequalities:

1)a1pa2k+a3k+⋯+atk+a2pa3k+a4k+⋯+at+1k+⋯+an−1pank+a1k+⋯+at−2k++anpa1k+a2k+⋯+at−1k≥(a1p+a2p+⋯+anp)2(t−1)(a1m+a2m+⋯+anm) 1) \quad \frac{a_1^p}{a_2^k + a_3^k + \cdots + a_t^k} + \frac{a_2^p}{a_3^k + a_4^k + \cdots + a_{t+1}^k} + \cdots + \frac{a_{n-1}^p}{a_n^k + a_1^k + \cdots + a_{t-2}^k} + \\ \qquad + \frac{a_n^p}{a_1^k + a_2^k + \cdots + a_{t-1}^k} \ge \frac{(a_1^p + a_2^p + \cdots + a_n^p)^2}{(t-1)(a_1^m + a_2^m + \cdots + a_n^m)}

2)a2k+a3k+⋯+atka1p+a3k+a4k+⋯+at+1ka2p+⋯+ank+a1k+⋯+at−2kan−1p++a1k+a2k+⋯+at−1kanp≥(t−1)(a1k+a2k+⋯+ank)2a1m+a2m+⋯+anm 2) \quad \frac{a_2^k + a_3^k + \cdots + a_t^k}{a_1^p} + \frac{a_3^k + a_4^k + \cdots + a_{t+1}^k}{a_2^p} + \cdots + \frac{a_n^k + a_1^k + \cdots + a_{t-2}^k}{a_{n-1}^p} + \\ \qquad + \frac{a_1^k + a_2^k + \cdots + a_{t-1}^k}{a_n^p} \ge \frac{(t-1)(a_1^k + a_2^k + \cdots + a_n^k)^2}{a_1^m + a_2^m + \cdots + a_n^m}

Solution

1) By the rearrangement inequality we have:
a 1 m + a 2 m + + a n m a 1 p a 2 k + a 2 p a 3 k + + a n-1 p a n k + a n p a 1 k a 1 m + a 2 m + + a n m a 1 p a 3 k + a 2 p a 4 k + + a n-1 p a 1 k + a n p a 2 k ................... ................... a 1 m + a 2 m + + a n m a 1 p a t k + a 2 p a t+1 k + + a n-1 p a t-2 k + a n p a t-1 k\text{a 1 m + a 2 m + + a n m a 1 p a 2 k + a 2 p a 3 k + + a n-1 p a n k + a n p a 1 k a 1 m + a 2 m + + a n m a 1 p a 3 k + a 2 p a 4 k + + a n-1 p a 1 k + a n p a 2 k ................... ................... a 1 m + a 2 m + + a n m a 1 p a t k + a 2 p a t+1 k + + a n-1 p a t-2 k + a n p a t-1 k}
Adding above (t−1)(t-1) inequalities we get
(1)(t−1)(a1m+a2m+⋯+anm)≥a1p(a2k+a3k+⋯+atk)+a2p(a3k+a4k+⋯+at+1k)+⋯+anp(a1k+a2k+⋯+at−1k) (1) \quad (t-1)(a_1^m + a_2^m + \dots + a_n^m) \ge a_1^p (a_2^k + a_3^k + \dots + a_t^k) + a_2^p (a_3^k + a_4^k + \dots + a_{t+1}^k) + \dots + a_n^p (a_1^k + a_2^k + \dots + a_{t-1}^k)

and using Cauchy-Schwartz inequality we have
(t−1)(a1m+a2m+⋯+anm)(a1pa2k+a3k+⋯+atk+a2pa3k+a4k+⋯+at+1k+⋯+anpa1k+a2k+⋯+at−1k)≥(a1p+a2p+⋯+anp)2 (t-1)(a_1^m + a_2^m + \dots + a_n^m) \left( \frac{a_1^p}{a_2^k + a_3^k + \dots + a_t^k} + \frac{a_2^p}{a_3^k + a_4^k + \dots + a_{t+1}^k} + \dots + \frac{a_n^p}{a_1^k + a_2^k + \dots + a_{t-1}^k} \right) \ge (a_1^p + a_2^p + \dots + a_n^p)^2
Now, our conclusion is obtained by dividing both sides of above equation by (t−1)(a1m+a2m+⋯+anm)(t-1)(a_1^m + a_2^m + \dots + a_n^m).

2) Multiplying both sides of (1) by
a2k+a3k+⋯+atka1p+a3k+a4k+⋯+at+1ka2p+⋯+a1k+a2k+⋯+at−1kanp \frac{a_2^k + a_3^k + \dots + a_t^k}{a_1^p} + \frac{a_3^k + a_4^k + \dots + a_{t+1}^k}{a_2^p} + \dots + \frac{a_1^k + a_2^k + \dots + a_{t-1}^k}{a_n^p}
and using Cauchy-Schwartz inequality too we have
(t−1)(a1m+a2m+⋯+anm)(a2k+a3k+⋯+atka1p+a3k+a4k+⋯+at+1ka2p+⋯+a1k+a2k+⋯+at−1kanp)≥[(t−1)(a1k+a2k+⋯+ank)]2 (t-1)(a_1^m + a_2^m + \dots + a_n^m) \left( \frac{a_2^k + a_3^k + \dots + a_t^k}{a_1^p} + \frac{a_3^k + a_4^k + \dots + a_{t+1}^k}{a_2^p} + \dots + \frac{a_1^k + a_2^k + \dots + a_{t-1}^k}{a_n^p} \right) \ge [(t-1)(a_1^k + a_2^k + \dots + a_n^k)]^2
Again, our conclusion is obtained by dividing both sides of above equation by (t−1)(a1m+a2m+⋯+anm)(t-1)(a_1^m + a_2^m + \dots + a_n^m).

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.