n variable nice inequality

[Corei13]

Prove that, for $a_1, a_2, \cdots a_n \geq 0$,
\[ \left( a_{1}+\frac{a_{2}}{2}+\cdots+\frac{a_{n}}{n}\right) (a_{1}+2 a_{2}+\cdots+n a_{n})\leq \frac{(n+1)^{2}}{4n} (a_{1}+a_{2}+\cdots+a_{n})^{2} \]

[sourav das]

By A.M-G.M
\[ \frac{1}{4n}\left ( 4\left ( \sum_{i=1}^{n}\frac{na_{i}}{i} \right ) \left ( \sum_{i=1}^{n}ia_{i} \right )\right )\leq \frac{1}{4n}\left ( \sum_{i=1}^{n}a_i\left ( \frac{n}{i}+i \right ) \right )^{2}\]
But by re-arrangement
\[\left ( \frac{n}{i}+i \right )\leq \left ( n+1 \right )\]

It implies:
\[\left( \sum_{i=1}^{n}\frac{1}{i}a_{i}\right )\left ( \sum_{i=1}^{n}ia_{i} \right )
\leq \frac{(n+1)^{2}}{4n}\sum_{i=1}^{n}a_{i}\]


My impression until i solve it:
But after cracking the solution it is really nice indeed.

[Corei13]

Generalization : ( Reversed Cauchy ? )
\[\left(\sum_{1\leq i \leq n }{a_i^2}\right)\left(\sum_{1\leq i \leq n }{b_i^2}\right) \leq \frac{(M+m)^2}{4Mm}\left(\sum_{1\leq i \leq n }{a_i b_i}\right)^2 \]
Where, $M=\max_{1\leq i \leq n }{\left\{\frac{a_i}{b_i} \right\}}$ and $m=\min_{1\leq i \leq n }{\left\{\frac{a_i}{b_i} \right \}}$

[sourav das]

I think it is defined for all positive reals

[Corei13]

Hmm, Yes.
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[sourav das]

SUPERRRRRR,

Define: $m_i= \frac{a_i}{b_i}$
So our given inequality transforms into:
\[\left ( \sum_{i=1}^{n}m_{i}a_{i}b_{i} \right )\left ( \sum_{i=1}^{n}\frac{1}{m_{i}}a_{i}b_{i} \right )\leq \left ( \frac{(M+m)^{2}}{4Mm} \right )\left ( \sum_{i=1}^{n}a_{i}b_{i} \right )\]
Now,by A.M-G.M
\[\frac{1}{4mM}4\left ( \sum_{i=1}^{n}m_{i}a_{i}b_{i} \right )\left ( \sum_{i=1}^{n}\frac{mM}{m_{i}}a_{i}b_{i} \right )\leq \frac{1}{4mM}\left ( \sum_{i=1}^{n}a_{i}b_{i}\left ( m_{i}+\frac{mM}{m_{i}} \right ) \right )\]....(i)
But by re-arrangement inequality, As, $m_{i} \geq m$ ; $M \geq m_{i}$
\[m_{i}^{2}+Mm\leq m_{i}m +m_{i}M\]
\[\Rightarrow m_{i}+\frac{Mm}{m_{i}}\leq m +M\]
Using this to (i) we are done.

Where did you get this? SUPER COOLLLLLLL

[Corei13]

http://www.artofproblemsolving.com/Wiki ... Inequality

[Corei13]

http://www.artofproblemsolving.com/Foru ... 4&sr=posts
A walk through Functional Equation!