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.

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.

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.

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.