Inegalitate conditionata cu suma de patrate (2)

[Marius Mainea]

Daca \( n\ge 4 \) esta natural si \( a_1,a_2,...,a_n \) pozitive, astfel incat \( a_1^2+a_2^2+...+a_n^2=1 \), atunci

a) \( \frac{1}{4}\ge a_1^2a_2^2+a_2^2a_3^2+...+a_{n-1}^2a_n^2+a_n^2a_1^2 \)

b) \( \frac{a_1}{a_2^2+1}+\frac{a_2}{a_3^2+1}+...+\frac{a_n}{a_1^2+1}\ge\frac{4}{5}(a_1\sqrt{a_1}+a_2\sqrt{a_2}+...+a_n\sqrt{a_n})^2
\)

[Marius Mainea]

Indicatie :

a) Se arata prin inductie ca pentru \( n\ge 4 \) avem \( (x_1+x_2+...+x_n)^2\ge 4(x_1x_2+x_2x_3+...+x_nx_1) \).

b) Se foloseste a).

[Mateescu Constantin]

b) \( a_1^2+a_2^2+...+a_n^2=1\ (\ast) \)

\( LHS=\sum_{k=1}^n {\frac{a_k^3}{a_k^2a_{k+1}^2+a_k^2}}\ \ge^{\mbox{C.B.S.}}\ \frac{\left(\sum_{k=1}^n{a_k\sqrt{a_k}}\right\)^2}{\sum_{k=1}^n{\(a_k^2a_{k+1}^2+a_k^2\)}}\ =^{(\ast)}\ \frac{\left\(\sum_{k=1}^n a_k\sqrt{a_k}\right\)^2}{1+\sum_{k=1}^n a_k^2a_{k+1}^2}\ \ge^{\mbox{a)}}\ \frac{\(a_1\sqrt{a_1}+a_2\sqrt{a_2}+...+a_n\sqrt{a_n}\)^2}{1+\frac 14}=RHS \)