A geometric matrix inequality Let $0 < \sigma_i \in \mathbb{R}$ ($i=1 \dots d$). Is it true that
$$ c \sum_{i=1}^d (\sigma_i-1)^2 \le \sqrt{\sum_{i=1}^d (\sigma_i^2-1)^2} \tag{1}$$
for some $c>0$ which does not depend on the $\sigma_i$.
(I actually suspect this holds for $c=1$).
Motivation:
This inequality is equivalent to the following
$$c \|\sqrt{A^TA}-I\|^2 \le \|A^TA-I\|, \tag{2}$$
where $A$ is a $d \times d$ real matrix. (The equivalence is obtained by considering SVD).
(Inequality $(2)$ comes from comparing different ways to measure deviation of a linear transformation from being an isometry).
 A: Isn't this just a direct application of Cauchy-Schwarz inequality? We have
\begin{align*}
\sum_{i=1}^d (\sigma_i-1)^2
&= \left\langle\left((\sigma_1-1)^2,\ldots,(\sigma_d-1)^2\right),\ (1,\ldots,1)\right\rangle\\
&\le \left\|\left((\sigma_1-1)^2,\ldots,(\sigma_d-1)^2\right)\right\|\cdot \|(1,\ldots,1)\|\\
&= \sqrt{d}\sqrt{\sum_i(\sigma_i-1)^4}\\
&= \sqrt{d}\sqrt{\sum_i\left[(\sigma_i-1)^4-(\sigma_i^2-1)^2+(\sigma_i^2-1)^2\right]}\\
&= \sqrt{d}\sqrt{\sum_i\left[-4\sigma_i^2(\sigma_i-1)^2+(\sigma_i^2-1)^2\right]}\\
&\le \sqrt{d}\sqrt{\sum_i(\sigma_i^2-1)^2}.
\end{align*}
Equality holds in the first inequality iff all $|\sigma_i-1|$s are equal, and equality holds in the second inequality iff each $\sigma_i$ is $0$ or $1$. Hence ties occur in both inequalities iff all $\sigma_i$ are equal to zero or all $\sigma_i$ are equal to one, i.e. iff $A=0$ or $A$ is real orthogonal.
A: Sorry for all the edits...
The inequality holds for $c=1/d$. Indeed,
$$
\sum(\sigma_i-1)^2 \le d \max_i (\sigma_i-1)^2 \le d \max_i |\sigma_i^2-1|\le d\sqrt{\sum (\sigma_i^2-1)^2},
$$
The inequality does not hold for any $c>1/\sqrt{d}$:
Assume $\sigma_i = n+1$ for all $i$. Then the inequality reads
$$ c\,d\,n^2 \le \sqrt{d(n^2+2n)^2} = \sqrt{d}(n^2+2n),$$
and by taking $n\to \infty$ you get that $c\le 1/\sqrt{d}$.
In fact, if $\sum \sigma_i \ge d$, then inequality holds with $c=1/\sqrt{d}$. Indeed, using Hölder's inequality, we have
$$
\sum(\sigma_i-1)^2 = \sum(\sigma_i^2-1) - 2\sum(\sigma_i-1) \le \sqrt{d}\sqrt{\sum (\sigma_i^2-1)^2} - 2\sum(\sigma_i-1).
$$
I suspect $c=1/\sqrt{d}$ is the best constant.
A: It's clear that in the case of $\sum_i(\sigma_i-1)^2 \leq 1$, we have
$$
\sqrt{\sum_{i=1}^d (\sigma_i^2-1)^2} = 
\sqrt{\sum_{i=1}^d (\sigma_i-1)^2(\sigma_i+1)^2} \geq 
\sqrt{\sum_{i=1}^d (\sigma_i-1)^2} \geq \sum_{i=1}^d (\sigma_i-1)^2
$$
I'm not sure about the general case, though.

Consider the case of $\sigma_1 = x+1 >\sigma_2 = y+1 > 1$ $\sigma_3 = \cdots = \sigma_d = 1$. We define
$$
f(x,y) = \sqrt{\sum_{i=1}^d (\sigma_i^2-1)^2} = \sqrt{[(x+1)^2 - 1]^2 + [(y+1)^2-1]^2}\\
= \sqrt{(x^2 + 2x)^2 + (y^2 + 2y)^2}\\
g(x,y) = \sum_{i=1}^d(\sigma_i - 1)^2 = x^2 + y^2
$$
Now, if we select $x = \alpha y^2$ for $\alpha > 0$, I think we'll notice something. I have a hunch that $f(x,y)/g(x,y)$ has no lower bound.
