18
$\begingroup$

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.

$\endgroup$
19
$\begingroup$

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

$\endgroup$
7
$\begingroup$

I have an other solution, which not as slick Markus Sprecher's one, but I think the method, which is quite general, is interesting in itself.

Fix $c \in \left]-1,1\right[$ and $n \geq 2$, and consider the compact manifold $$ \Sigma = \{ (a,b) \in S^{n-1} \times S^{n-1} \ | \ \langle a , b \rangle = c \}. $$ It has dimension $2n - 3$, and the tangent space at $(a,b)$ is the orthogonal in $\mathbb{R}^{2n}$ to the 3-space spanned by $(a,0),(0,b),(b,a)$.

Consider the functions $$ r_1 = \sqrt{\sum_i a_i^4}, \ \ r_2 = \sqrt{\sum_i b_i^4}, \ \ r_3 = \sum_{i} a_i^2 b_i^2 $$ as continuous functions on $\Sigma$. Let $(a,b)$ be a point of $\Sigma$ where $f = r_1 r_2 + r_3$ is maximal. We would like to show that $f(a,b) \leq 1 + c^2$.

The gradient of $f$ at $(a,b)$ must be of the form $\alpha(a,0) + \beta(0,b) + \gamma(b,a)$ for some real numbers $\alpha,\beta,\gamma$. This gives the following $2n$ equations: $$ (1) \ \ a_i b_i^2 + a_i^3 \frac{r_2}{r_1} = \alpha a_i + \gamma b_i \\ (2) \ \ b_i a_i^2 + b_i^3 \frac{r_1}{r_2} = \beta b_i + \gamma a_i \\ $$ Multiplying $(1)$ by $a_i$ and summing ober $i$, we get $f = r_3 +r_1r_2 = \alpha + c \gamma$. Multiplying $(2)$ by $b_i$ and summing ober $i$, we get $f = r_3 +r_1r_2 = \beta + c \gamma$. In particular $\alpha = \beta$.

Mutliply equations $(1)$ and $(2)$ by $r_1 b_i$ and $r_2 a_i$ respectively. This yields $$ (1)' \ \ r_1 a_i b_i^3 + r_2 a_i^3 b_i = r_1 \alpha a_i b_i + r_1 \gamma b_i^2 \\ (2)' \ \ r_2 a_i^3 b_i + r_1 a_i b_i^3 = r_2 \alpha a_ib_i + r_2 \gamma a_i^2 \\ $$ The two LHSs are equal, hence so are the RHSs. Thus $(a_i,b_i)$ satisifies the quadratic equation $$ (3) r_1 \alpha a_i b_i + r_1 \gamma b_i^2 = r_2 \alpha a_ib_i + r_2 \gamma a_i^2. $$

There are two cases:

  • if $\gamma =0$, then $\alpha = f >0$, and thus $r_1 = r_2$ follows from $(3)$. Then $(1)-(2)$ become the alternative $(a_i,b_i) =0$ or $a_i^2+ b_i^2 = \alpha$. Since $a$ and $b$ are not proportional, the latter cas happens at least twice, and thus $2 = ||a||_2^2 + ||b||_2^2 \geq 2 \alpha$. Thus $f = \alpha \leq 1$, and in particular $f \leq 1 + c^2$.
  • if $\gamma \neq 0$ then there are at most $2$ projective solutions to $(3)$. Since the system $(1)-(2)$ is inhomogeneous, it has at most $2$ non-zero solutions up to sign. Thus there are non zero vectors $(a_1'',b_1'')$ and $(a_2'',b_2'')$ such that for eachi $i$ one has either $(a_i,b_i) = \pm (a_j'',b_j'')$ for some $j=1,2$, or $(a_i,b_i) = 0$. Let $[|1,n|] = I_0 \sqcup I_1 \sqcup I_2$ such that $(a_i,b_i) = 0$ for $i$ in $I_0$, and $(a_i,b_i) = \pm (a_j'',b_j'')$ if $i$ is in $I_j$ for some $j=1,2$. We then have the identities \begin{align*} |I_1| (a_1'')^2 + |I_2| (a_2'')^2 &= \sum_i a_i^2 = 1 \\ |I_1| a_1''b_1'' + |I_2| a_2'' b_2'' &= \sum_i a_i b_i = c \\ |I_1| (b_1'')^2 + |I_2| (b_2'')^2 &= \sum_i b_i^2 = 1 \end{align*} Thus the vectors $a' = (|I_1|^{\frac{1}{2}} a_1'',|I_2|^{\frac{1}{2}}a_2'')$ and $b' = (|I_1|^{\frac{1}{2}} b_1'',|I_2|^{\frac{1}{2}}b_2'')$ in $\mathbb{R}^2$ satisfy $$ ||a'||_2 =1, \ \ ||b'|| = 1, \ \ \langle a',b' \rangle = c. $$ Moreover, one has \begin{align*} \sum_{i=1}^2 (a_i')^4 &= |I_1|^2 (a_1'')^4 + |I_2|^2 (a_2'')^4 \geq |I_1| (a_1'')^4 + |I_2| (a_2'')^4 = r_1^2 \\ \sum_{i=1}^2 (b_i')^4 &= |I_1|^2 (b_1'')^4 + |I_2|^2 (b_2'')^4 \geq |I_1| (b_1'')^4 + |I_2| (b_2'')^4 = r_2^2 \\ \sum_{i=1}^2 (a_i')^2(b_i')^2 &= |I_1|^2 (a_1'')^2(b_1'')^2 + |I_2|^2 (a_2'')^2(b_2'')^2 \geq |I_1| (a_1'')^2(b_1'')^2 + |I_2| (a_2'')^2 (b_2'')^2 = r_3 \\ \end{align*} Consequently, applying the case $n=2$ of the inequality to the vectors $a',b'$ yields $$ 1 + c^2 \geq r_1 r_2 + r_3 = f(a,b). $$ Thus $f \leq 1+c^2$ on all of $\Sigma$.
$\endgroup$
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Well, you claimed the equivalence between the two inequalities in a comment of that question. However, this equivalence has not been completely clear to me. That’s why I posted this question as a “conjecture” :) $\endgroup$ – Ludwig Jan 18 '17 at 18:44
  • $\begingroup$ Now I see, thanks! I edited my question in order to acknowledge this. $\endgroup$ – Ludwig Jan 19 '17 at 10:01
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I can't quite see why "After expansion, the LHS of (2) is a polynomial in even powers of $x$ and $y$." What about the $2xyp_1(ab)$ term in $(xy+p_1(ab))^2$, which is then cross multiplied by all the other even-power terms? $\endgroup$ – Fan Zheng Jan 16 '17 at 21:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.