Let $a\_i,b\_i\in\mathbb{R}$ and $n>1$, does the inequality 
$$
\left(\sum\_{i=1}^n a\_i^2\right)\left(\sum\_{i=1}^n b\_i^2\right)+\left(\sum\_{i=1}^na\_i b\_i\right)^2\ge \sqrt{\left(\sum\_{i=1}^n a\_i^4\right)\left(\sum\_{i=1}^n b\_i^4\right)}+\sum\_{i=1}^na\_i^2b\_i^2
$$
hold true?


Observe that $a\_i$ and $b\_i$ are **not** required to be non-negative. 
I ran an extensive number of numerical simulations and no counterexample showed up yet.


**Note 1.** The inequality holds true for $n=2$, as showed [here](http://www.wolframalpha.com/input/?i=((a_1%5E2%2Ba_2%5E2)(b_1%5E2%2Bb_2%5E2)%2B(a_1*b_1%2Ba_2*b_2)%5E2-(a_1%5E2*b_1%5E2%2Ba_2%5E2*b_2%5E2))%5E2-(a_1%5E4%2Ba_2%5E4)(b_1%5E4%2Bb_2%5E4)).


**Note 2.** This conjecture was formulated by Fedor Petrov in an attempt to provide a solution to a particular case of [this question](https://mathoverflow.net/questions/233546/a-log-inequality-for-positive-definite-trace-one-matrices).


**Note 3.** I have posted this question on [math.SE](https://math.stackexchange.com/questions/1706416/sum-limits-i-a-i2-sum-limits-i-b-i2-left-sum-limits-ia-i-b-i-right2-geq) some time ago but it has received no answer, so I cross-posted it here.


**Note 4.** As Fedor Petrov rightly observed in his answer below, the inequality also follows by an argument used in one of his answers in the above-cited question. However, I decided to accept Markus Sprecher's answer because of its conciseness and clarity.



A seemingly simple inequality
The inequality is equivalent to
$$
\left(\sum\_{i>j} (a\_ib\_j+a\_jb\_i)^2+\sum\_{i} a^2\_ib^2\_i \right)^2\geq \sum\_{i} a^4\_i \sum\_{i} b^4\_i.
$$
The left hand side is greater or equal to
$$
\sum\_i a\_i^4b\_i^4+\sum\_{i>j} (a\_ib\_j+a\_jb\_i)^4+2(a\_ib\_j+a\_jb\_i)^2(a\_i^2b\_i^2+a\_j^2b\_j^2)+2a\_i^2b\_i^2a\_j^2b\_j^2
$$
As
$$
(a\_ib\_j+a\_jb\_i)^4+2(a\_ib\_j+a\_jb\_i)^2(a\_i^2b\_i^2+a\_j^2b\_j^2)+2a\_i^2b\_i^2a\_j^2b\_j^2\geq a\_i^4b\_j^4+a\_j^4b\_i^4.
$$
is equivalent to
$$
(a\_ib\_j+a\_jb\_i)^2(a\_ib\_i+a\_jb\_j)^2\geq 0
$$
the LHS is larger or equal to
$$
\sum\_i a\_i^4b\_i^4+\sum\_{i>j} a\_i^4b\_j^4+a\_j^4b\_i^4=\sum\_{i} a^4\_i \sum\_{i} b^4\_i.
$$


