Given $\sum\limits_{i=1}^6a_i^2=6$, where $a_i>0$, $a_7=a_1$. Prove that $\sum\limits_{i=1}^6\frac{a_i^2}{a_{i+1}}\geq6$ Let $a_i$ be positive numbers such that $a_1^2+a_2^2+a_3^2+a_4^2+a_5^2+a_6^2=6$. Prove that:
$$\frac{a_1^2}{a_2}+\frac{a_2^2}{a_3}+\frac{a_3^2}{a_4}+\frac{a_4^2}{a_5}+\frac{a_5^2}{a_6}+\frac{a_6^2}{a_1}\geq6$$
I tried C-S and SOS, but without success. 
 A: We can write:
$$ \frac{a_1^2}{a_2} + \frac{a_2^2}{a_2} + \ldots + \frac{a_6^2}{a_1} \geq
a_1^2 + \ldots + a_6^2$$ 
which is equivalent to this inequality.
after dividing both sides by $(a_1^2 + \ldots + a_6^2)$ we get:
$$ \frac{a_1^2}{a_1^2 + \ldots + a_6^2} \frac{1}{a_2} +  \ldots +   \frac{a_6^2}{a_1^2 + \ldots + a_6^2} \frac{1}{a_1}  \geq
1$$ 
Function $f(x)= \frac{1}{x}$ is convex in real positive domain, hence from Jensen's inequality we have:
$$\frac{a_1^2}{a_1^2 + \ldots + a_6^2} \frac{1}{a_2} +  \ldots +   \frac{a_6^2}{a_1^2 + \ldots + a_6^2} \frac{1}{a_1}
\geq
\frac{1}{ \frac {a_1^2 a_2 } {a_1^2 + \ldots + a_6^2} +
 \frac {a_2^2 a_3 } {a_1^2 + \ldots + a_6^2} +
 \ldots + \frac {a_6^2 a_1 } {a_1^2 + \ldots + a_6^2} }
$$.
$$\frac{a_1^2}{a_1^2 + \ldots + a_6^2} \frac{1}{a_2} +  \ldots +   \frac{a_6^2}{a_1^2 + \ldots + a_6^2} \frac{1}{a_1}
= \frac{a_1^2 a_2 + a_2^2 a_3 + \ldots + a_6^2 a_1 } {a_1^2 + \ldots + a_6^2}
$$
but 
$$ a_1^2 a_2 + \ldots + a_6^2 a_1 \geq a_1^2 + \ldots + a_6^2 $$
because by dividing both sides by $a_1^2 + \ldots + a_6^2$
we have
$$ \frac{a_1^2}{  a_1^2 + \ldots + a_6^2} a_2 + \ldots + \frac{a_6^2} {  a_1^2 + \ldots + a_6^2} a_1 \geq a_1+ \ldots + a_6 \geq 1.$$ (Jensen's inequality)
$$\frac{a_1^2 a_2 + a_2^2 a_3 + \ldots + a_6^2 a_1 } {a_1^2 + \ldots + a_6^2}\geq 1
$$
and finally:
$$ \frac{a_1^2}{a_1^2 + \ldots + a_6^2} \frac{1}{a_2} +  \ldots +   \frac{a_6^2}{a_1^2 + \ldots + a_6^2} \frac{1}{a_1}  \geq
1$$ is true.
