Click to reveal hidden text

n=3: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=497887&p=3121079#p3121079 here here

Let $a_{1}, a_{2}, \cdots, a_{n}$ be non-negative numbers such that $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=1$($n=100$). Prove that $$a_{1}^{2}a_{2}+a_{2}^{2}a_{3}+\cdots+a_{n}^{2}a_{1}< \frac{\sqrt{2}}{3}.$$

N.T.TUAN wrote: If $x_{1},x_{2},...,x_{n}(n>2)$ are positive real numbers with $x_{1}+x_{2}+...+x_{n}=1$. Prove that $x_{1}^{2}x_{2}+x_{2}^{2}x_{3}+...+x_{n}^{2}x_{1}\leq\frac{4}{27}$. Assume $x_{1}\ge {x_{i} },(i=2,3,4,...,n)$, $a=x_{1},b=x_{2},c=x_{3}+x_{4}+...+x_{n} \rightarrow a+b+c=1 , (a,b,c\ge 0)$ $$LHS\leq x^2_{1}x_{2}+x^2_{2}x_{3}+(x^2_{3}+x^2_{4}+...+x^2_{n})x_{1}$$$$LHS\leq x^2_{1}x_{2}+x^2_{2}(x_{3}+x_{4}+...+x_{n})+(x_{3}+x_{4}+...+x_{n})^2x_{1}$$$$LHS\leq a^2b+b^2c+c^2a\leq \frac{4}{27}(a+b+c)^3$$

2003, I proof : Let $x_{1},x_{2},...,x_{n}(n\geq 3)$ are non-negative numbers with $x_{1}+x_{2}+\cdots+x_{n}=1$. Prove that $$x_{1}^{2}x_{2}+x_{2}^{2}x_{3}+...+x_{n}^{2}x_{1} +x_{1} x_{2} \cdots x_{n}\leq\frac{4}{27}.$$

Attachments:

sqing wrote: Let $x_{1},x_{2},...,x_{n}(n\geq 3)$ are non-negative numbers with $x_{1}+x_{2}+\cdots+x_{n}=1$. Prove that $$x_{1}^{2}x_{2}+x_{2}^{2}x_{3}+...+x_{n}^{2}x_{1} +x_{1} x_{2} \cdots x_{n}\leq\frac{4}{27}.$$ because $0\leq x_{i}\leq 1 \rightarrow x_{1}x_{2}x_{3}...x_{n}\leq x_{1}x_{2}x_{3}\leq x_{1}x_{2}(x_{3}+x_{4}+...+x_{n})=abc$ $$\rightarrow LHS \leq a^2b+b^2c+c^2a+abc\leq \frac{4}{27}(a+b+c)^3$$

Let $a$, $b$ and $c$ be non-negatives. Prove that$$ab^2+bc^2+cd^2+da^2\leq\frac{4}{27}(a+b+c)^3.$$Let $d=\min\{a,b,c,d\}$.Thus,$$ab^2+bc^2+cd^2+da^2=ab^2+bc^2+ca^2+abc+cd^2-abc+da^2-ca^2$$$$\leq ab^2+bc^2+ca^2+abc\leq\frac{4}{27}(a+b+c)^3.$$