I have no idea about how to prove (or disprove) the following inequality: $$ \left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right)+\left(\sum_{i=1}^na_i b_i\right)^2\geq \sqrt{\left(\sum_{i=1}^n a_i^4\right)\left(\sum_{i=1}^n b_i^4\right)}+\sum_{i=1}^na_i^2b_i^2,\quad a_i,b_i\in\mathbb{R}, \ n>1. $$ I ran some numerical simulations and no counterexample showed up yet.

Note 1. The inequality holds true for $n=2$, as shown here.

Note 2. This conjecture was formulated by Fedor Petrov in an attempt to provide a solution to a particular case of this question.

EDIT. The inequality has been finally proved here.

  • $\begingroup$ What happens if you square both sides and try to match up terms? $\endgroup$ Mar 20 '16 at 22:38
  • $\begingroup$ @MichaelBurr: I tried but it didn't help me. Also, I think that Cauchy-Schwarz inequality can help at some point, but I'm still stuck at the moment... $\endgroup$
    – Ludwig
    Mar 21 '16 at 8:59
  • $\begingroup$ $a_i$, $b_i$ should be non-negative. $\endgroup$ Mar 22 '16 at 10:01
  • $\begingroup$ Where does it inequality come from? When $a_i,b_i$ have the same sign for any $i$, it's clearly true, but otherwise it's not easy. $\endgroup$
    – Khue
    Mar 23 '16 at 14:18
  • 1
    $\begingroup$ It may turn out as a crazy idea, but have you tried to prove the inequality by induction on $n$? $\endgroup$ Aug 11 '16 at 2:13

This answer comes almost entirely from Markus Sprecher on MathOverflow found here.

The inequality is equivalent to $$ \left(\sum_{i>j} (a_ib_j+a_jb_i)^2+\sum_{i=1}^n (a_ib_i)^2 \right)^2\geq \left(\sum_{i=1}^n a^4_i\right)\left( \sum_{i=1}^n b^4_i\right). $$ The left hand side is greater than or equal to $$ \sum_i a_i^4b_i^4+\sum_{i>j} (a_ib_j+a_jb_i)^4+2(a_ib_j+a_jb_i)^2\big((a_ib_i)^2+(a_jb_j)^2)+2(a_ib_i)^2\cdot(a_jb_j)^2 $$ Since $$ (a_ib_j+a_jb_i)^4+2(a_ib_j+a_jb_i)^2\big((a_ib_i)^2+(a_jb_j)^2\big)+2(a_ib_i)^2\cdot(a_jb_j)^2\geq a_i^4b_j^4+a_j^4b_i^4. $$ is equivalent to $$ (a_ib_j+a_jb_i)^2(a_ib_i+a_jb_j)^2\geq 0, $$ the LHS is greater than or equal to $$ \sum_{i=1}^n a_i^4b_i^4+\sum_{i>j} a_i^4b_j^4+a_j^4b_i^4=\sum_{i=1}^n a^4_i \sum_{i=1}^n b^4_i. $$

  • 1
    $\begingroup$ I added the answer (with minor revisions) to clear this question from the unanswered queue following the suggestion of Henning found here. If it should be marked as CW, please mark it as such. $\endgroup$
    – Clayton
    Feb 27 '18 at 15:38

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