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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.

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.

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.

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.

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.

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.