# 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.