Let nonnegative real numbers $ a_{1},a_{2},a_{3},a_{4}$ satisfy $ a_{1} + a_{2} + a_{3} + a_{4} = 1.$ Prove that $ max\{\sum_{1}^4{\sqrt {a_{i}^2 + a_{i}a_{i - 1} + a_{i - 1}^2 + a_{i - 1}a_{i - 2}}},\sum_{1}^4{\sqrt {a_{i}^2 + a_{i}a_{i + 1} + a_{i + 1}^2 + a_{i + 1}a_{i + 2}}}\}\ge 2.$ Where for all integers $ i, a_{i + 4} = a_{i}$ holds.
Problem
Source: Chinese TST 2009 5th P3
Tags: inequalities, inequalities proposed
hxy09
11.04.2009 18:37
Do any people want to have a try?This is the unique small-inequality in China TST.It looks terrible,right?But it indeed have a two line proof,try and solve it mathlinkers p.s. I failed to solve it in the exam,but when I see the two line proof,I was wondering how stupid I am.
linboll
12.04.2009 09:35
This is the solution I came up with during the test.maybe it is the "two lines solution" Lemma:for non-negative real numbers a,b,c, $ \sqrt {b^2+a^2+ba+bc}+\sqrt {b^2+c^2+ba+bc} \ge 2b+a+c$ Proof: square both sides, it is equivalent to $ {(b^2+a^2+ba+bc)}*{(b^2+c^2+ba+bc)} \ge {(b^2+ac+ba+bc)}^2$, which is just an application of the famous Cauchy Inequality. Back to the problem, take the sum of the two terms in the blanket.It suffices to prove the sum is not less than 4.Using the lemma four times we get the sum is not less than $ 4(a_{1}+a_{2}+a_{3}+a_{4})=4$. Q.E.D
hxy09
12.04.2009 11:07
You're right ! p.s. congratulate you on being selected to be the member of our national team,wish you a perfect performance on 2009 IMO
linboll
12.04.2009 16:18
thanks PS: what's your name?could you tell me via short message?
hxy09
13.04.2009 12:10
linboll wrote: thanks PS: what's your name?could you tell me via short message? Thank you for attention,and I have sent you a message,please check your inbox