Proof or disproof $Y = 2\sum_{i=1}^{n} a^{2}_{i} b^{2}_{i} + \sum_{i \not=j}^{n} a^{2}_{i}b^{2}_{j} - 2\sum_{i=1}^{n} a_{i} b_{i} \ge 0$

Question

In my attempt to answer this question

I came cross this question :

if $$ \sum_{i=1}^{n}a^2_{i}\le 1,\sum_{i=1}^{n}b^2_{i}\le 1 $$ do we have ?

$$Y = 2\sum_{i=1}^{n} a^{2}_{i} b^{2}_{i} + \sum_{i \not=j}^{n} a^{2}_{i}b^{2}_{j} - 2\sum_{i=1}^{n} a_{i} b_{i} \ge 0$$

This obviously fails for $n=1$. – PA6OTA Jun 20 '14 at 16:09 — PA6OTA, Jun 20 '14 at 16:09
and for $n > 1$ ? – Naoufal EL JAOUHARI Jun 20 '14 at 16:16 — Naoufal EL JAOUHARI, Jun 20 '14 at 16:16
Hello, Note $a_{i}>0,b_{i}>0$ – math110 Jun 20 '14 at 17:23 — math110, Jun 20 '14 at 17:23
No, this inequelity is false .., your inequality is a rock. – Naoufal EL JAOUHARI Jun 20 '14 at 17:30 — Naoufal EL JAOUHARI, Jun 20 '14 at 17:30

score 0 · Accepted Answer · answered Jun 20 '14 at 16:30

0

Take a=(0.5 , 0) and b=(1 , 0) to show the inequality is wrong for n=2.

Conclude by recurring (adding 0 to each vector last coordinate).

answered Jun 20 '14 at 16:30

Colonel Beauvel

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