A seemingly simple inequality 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.
Note 2. This conjecture was formulated by Fedor Petrov in an attempt to provide a solution to a particular case of this question.
Note 3. I have posted this question on math.SE 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: Just to clarify. As I remember, this is simply equivalent to the (partial case of) the question you cite. Since the cited question was solved, I would not call it a conjecture. But the question to find an independent proof makes sense. 
Well, let me elaborate. Here proving the partial case $N=2$ of your inequality I prove the following inequality: $$\frac{\|x\|^2\cdot \|y\|^2+(x,y)^2}{\|Tx\|^2 \|Ty\|^2}\geqslant \frac2{{\rm tr}\, T^4}$$
for any self-adjoint positive definite operator $T$ and any two vectors $x,y$ in $\mathbb{R}^n$. If we denote $x=(a_1,\dots,a_n)$, $y=(b_1,\dots,b_n)$, $p=(a_1^2,\dots,a_n^2)$, $q=(b_1^2,\dots,b_n^2)$, $T^2=diag(s_1,\dots,s_n)$, $s=(s_1,\dots,s_n)$, we rewrite this as $$\left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right)+\left(\sum_{i=1}^na_i b_i\right)^2\ge 2\frac{(p,s)\cdot (q,s)}{(s,s)},$$
and maximizing over $s$ gives you  $$\sqrt{\left(\sum_{i=1}^n a_i^4\right)\left(\sum_{i=1}^n b_i^4\right)}+\sum_{i=1}^na_i^2b_i^2$$ in RHS, this is the equality case in the lemma in the cited answer.
A: Denote the power-sum (symmetric) polynomials by $p_k(a;n)=\sum_{i=1}^na_i^k$; where $a=(a_1,\dots, a_n)$ and also we need the Hadamard product $ab=(a_1b_1,\dots,a_nb_n)$. Write $p_k(a)$ for $p_k(a;n)$ when there is no confusion. 
Remark. Observe that $p_2(a)p_2(b)-p_2(ab)=\sum_{i\neq j}a_i^2b_j^2\geq0$. We prove
$$[p_2(a)p_2(b)+p_1(ab)^2-p_2(ab)]^2-p_2(a^2)p_2(b^2)\geq0.\tag1$$
Induct on $n$. The case $n=2$ is noted as trivial. Assume true for $n$, we show for $n+1$; i.e.
$$[(x^2+p_2(a))(y^2+p_2(b))+(xy+p_1(ab))^2-x^2y^2-p_2(ab)]^2-(x^4+p_2(a^2))(y^4+p_2(b^2))\geq0.\tag2$$
After expansion, the LHS of (2) is a polynomial in even powers of $x$ and $y$. It suffices to study the following coefficients (the others follow by symmetry):
$[x^0y^0]$: the is exactly the induction assumption, hence positive.
$[x^2y^0]$: $2p_2(b)[p_1(ab)^2+p_2(a)p_2(b)-p_2(ab)]\geq0$, by remark from above.
$x^4y^0]$: $p_2(b)^2-p_2(b^2)\geq0$, by remark.
$[x^2y^2]$: $6p_1(ab)^2+4p_2(a)p_2(b)-2p_2(ab)\geq0$, by remark.
$[x^4y^2]$: $2p_2(b)\geq0$.
$[x^0y^4]$: $p_2(a)^2-p_2(a^2)\geq0$, by remark.
Therefore, the claim (1) is valid for all $n$.
The intent of this method is to reveal that the inequality (1) is not as sharp as we may wish.
A: 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
$$
\sum_i a_i^4b_i^4+\sum_{i>j} a_i^4b_j^4+a_j^4b_i^4=\sum_{i} a^4_i \sum_{i} b^4_i.
$$
