# 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.

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

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 $\Sigma$.

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.

• 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” :) – Ludwig Jan 18 '17 at 18:44
• Now I see, thanks! I edited my question in order to acknowledge this. – Ludwig Jan 19 '17 at 10:01

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.

• 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? – Fan Zheng Jan 16 '17 at 21:13