2007 IMO SL #A6

Let $ a_1, a_2, \ldots, a_{100}$ be nonnegative real numbers such that $ a^2_1 + a^2_2 + \ldots + a^2_{100} = 1.$ Prove that \[ a^2_1 \cdot a_2 + a^2_2 \cdot a_3 + \ldots + a^2_{100} \cdot a_1 < \frac {12}{25}. \]

By the Cauchy-Schwartz inequality, note that $$\frac{1}{3}[a_1(a_{100}^2+2a_1a_2)+a_2(a_1^2+2a_2a_3)+\dots+a_{100}(a_99^2+2a_{100}a_1)]$$ $$\le \frac{1}{3}(a_1^2+a_2^2+a_3^2+\dots+a_{100}^2)^{\frac{1}{2}} \cdot \left(\sum_{k=1}^{100}(a_k^2+2a_{k+1}a_{k+2})\right)^{\frac{1}{2}},$$ in which the indices are modulo $100$. 

Therefore, it suffices to prove $$\sum_{k=1}^{100}(a_k^2+2a_{k+1}a_{k+2})^2 \le 2.$$ Each and every term of the sum $\sum_{k=1}^{100}(a_k^2+2a_{k+1}a_{k+2})^2$ can be written as $$a_k^4+4a_{k+1}^2a_{k+2}^2+4a_k^2(a_{k+1} \cdot a_{k+2}) \le (a_k^4+2a_k^2a_{k+1}^2+2a_k^2a_{k+2})+4a_{k+1}^2a_{k+2}^2,$$ which is true because

$$2(a_{k+1}\cdot a_{k+2}) \le (a_{k+1}^2+a_{k+2}^2).$$Finally, our desired inequality follows from adding the inequalities $$\sum_{k=1}^{100}(a_k^4+2a_k^2a_{k+1}^2+2a_k^2a_{k+2}^2) \le (a_1^2+\cdots+a_{100}^2)^2=1$$ and $$\sum_{k=1}^{100}a_k^2a_{k+1}^2 \le (a_1^2+a_3^2+\cdots+a_{99}^2) \cdot (a_2^2+a_4^2+\cdots+a_{100}^2) \le \frac{1}{4}(a_1^2+a_2^2+\cdots+a_{100}^2)^2 = \frac{1}{4}.$$

We are done. $\square$


  1. Anonymous1/06/2022

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