we have : $\Big((a_1+a_2)+(a_2+a_3)+...+(a_{n}+a_1)\Big)\Big(\frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+\dots+\frac{a_n^2}{a_n+a_1}\Big) \geq (a_1+a_2+...+a_n)^2$ (by caushy-scharws) So, we have : $\frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+\dots+\frac{a_n^2}{a_n+a_1} \geq \frac{1^2}{2 \cdot 1}=\frac{1}{2}$

Posted before: http://www.mathlinks.ro/Forum/viewtopic.php?t=5148

By generalized Titu's lemma LHS$\ge \frac{(\sum{a_i})^2}{2(\sum{a_i})}=\frac{\sum{a_i}}{2}=\frac{1}{2}$ Too easy for a TST, isn't it..... Well,my school exams are near,so I am satisfied with easy ones.... Bye... Sayantan....

Xixas wrote: Let $a_1, a_2, \dots, a_n$ be positive real numbers, whose sum is $1$. Prove that \[ \frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+\dots+\frac{a_{n-1}^2}{a_{n-1}+a_n}+\frac{a_n^2}{a_n+a_1}\ge \frac{1}{2} \] Let $a_{n+1} = a_{1}$. We have: $$\sum_{i=1}^{n} \frac{a_{i}^2}{a_{i}+a_{i+1}}=\frac{1}{2}\left(\sum_{i=1}^{n} \frac{a_{i}^2+a_{i+1}^2}{a_{i}+a_{i+1}}\right) \geq \frac{1}{4}\left(\sum_{i=1}^{n} (a_{i}+a_{i+1})\right) = \frac{1}{2} $$

\[ \frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+\dots+\frac{a_{n-1}^2}{a_{n-1}+a_n}+\frac{a_n^2}{a_n+a_1}\ge \frac{(a_1 + a_2 + a_3 + a_4 + a_5 + \cdots + a_n)^2}{2(a_1 + a_2 + a_3 + a_4 + a_5 + \cdots + a_n)} = \frac{1}{2}\] I have used Cauchy-Schwartz Lemma here.

dwip_neel wrote: \[ \frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+\dots+\frac{a_{n-1}^2}{a_{n-1}+a_n}+\frac{a_n^2}{a_n+a_1}\ge \frac{(a_1 + a_2 + a_3 + a_4 + a_5 + \cdots + a_n)^2}{2(a_1 + a_2 + a_3 + a_4 + a_5 + \cdots + a_n)} = \frac{1}{2}\] I have used Cauchy-Schwartz Lemma here. It's too easy as a TST problem.

$$\frac{a_1^2}{a_1+a_2}+\ldots+\frac{a_n^2}{a_n+a_1}\overset{\text{T2}}\ge\frac{(a_1+\ldots+a_n)^2}{2(a_1+\ldots+a_n)}=\frac12$$

By Titu's lemma: $\frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+\dots+\frac{a_{n-1}^2}{a_{n-1}+a_n}+\frac{a_n^2}{a_n+a_1} \ge \frac{(a_1+a_2+\dots+a_n)^2}{2(a_1+a_2+\dots +a_n)} = \frac{1}{2}$.$\blacksquare$

$$\sum_{i=1}^{n}\frac{a_i^2}{a_i+a_{i+1}}\overset{\text{Engel C-S}}{\ge}\frac{\left(\sum_{i=1}^{n}a_i\right)^2}{2\sum_{i=1}^{n}a_i}=\frac{1}{2}$$ $\blacksquare$

Xixas wrote: Let $a_1, a_2, \dots, a_n$ be positive real numbers, whose sum is $1$. Prove that \[ \frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+\dots+\frac{a_{n-1}^2}{a_{n-1}+a_n}+\frac{a_n^2}{a_n+a_1}\ge \frac{1}{2} \] More generally, if $a_1+\cdots +a_n=b_1+\cdots +b_n=1$ and $a_1,\ldots ,a_n,b_1,\ldots ,b_n>0$, then by the Fractionary form of Cauchy's Inequality we have$$\frac{a_1^2}{a_1+b_1}+\cdots +\frac{a_n^2}{a_n+b_n}\geq \frac{(a_1+\cdots +a_n)^2}{(a_1+\cdots +a_n)+(b_1+\cdots +b_n)}=\frac{1}{2}.$$We have equality iff $a_i=b_i$ for $i=1,\ldots ,n$, in the original problem this is $a_1=\cdots =a_n=\frac{1}{n}$.