$\sum\limits_i a_i^2\sum\limits_i b_i^2+\left(\sum\limits_ia_i b_i\right)^2\geq \sqrt{\sum\limits_i a_i^4\sum\limits_i b_i^4}+\sum\limits_ia_i^2b_i^2$

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.

• What happens if you square both sides and try to match up terms? – Michael Burr Mar 20 '16 at 22:38
• @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... – Ludwig Mar 21 '16 at 8:59
• $a_i$, $b_i$ should be non-negative. – Quang Hoang Mar 22 '16 at 10:01
• 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. – Khue Mar 23 '16 at 14:18
• It may turn out as a crazy idea, but have you tried to prove the inequality by induction on $n$? – Jack D'Aurizio Aug 11 '16 at 2:13

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 $$\sum_{i=1}^n a_i^4b_i^4+\sum_{i>j} a_i^4b_j^4+a_j^4b_i^4=\sum_{i=1}^n a^4_i \sum_{i=1}^n b^4_i.$$