A seemingly simple inequality | CodeUtility

The inequality is equivalent to
$$
left(sum_{i>j} (a_ib_j+a_jb_i)^2+sum_{i} a^2_ib^2_i right)^2geq sum_{i} a^4_i sum_{i} b^4_i.
$$
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
$$
(a_ib_j+a_jb_i)^4+2(a_ib_j+a_jb_i)^2(a_i^2b_i^2+a_j^2b_j^2)+2a_i^2b_i^2a_j^2b_j^2geq a_i^4b_j^4+a_j^4b_i^4.
$$
is equivalent to
$$
(a_ib_j+a_jb_i)^2(a_ib_i+a_jb_j)^2geq 0
$$
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 $c in left]-1,1right[$ 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: $$frac{|x|^2cdot |y|^2+(x,y)^2}{|Tx|^2 |Ty|^2}geqslant frac2{{rm tr}, T^4}$$
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 $$left(sum_{i=1}^n a_i^2right)left(sum_{i=1}^n b_i^2right)+left(sum_{i=1}^na_i b_iright)^2ge 2frac{(p,s)cdot (q,s)}{(s,s)},$$
and maximizing over $s$ gives you $$sqrt{left(sum_{i=1}^n a_i^4right)left(sum_{i=1}^n b_i^4right)}+sum_{i=1}^na_i^2b_i^2$$ in RHS, this is the equality case in the lemma in the cited answer.