Gaussian width vs. Gaussian complexity Source: Vershynin High Dimensional Probability page 181 Exercise 7.6.9
Consider the subset $T\subset \mathbb R^n$. Its Gaussian complexity is defined to be
$\gamma(T):=\mathbb{E}\sup_{x\in T} \left\vert \big<g, x\big> \right \vert $ and its Gaussian width is $w(T):=\mathbb{E}\sup_{x\in T} \big<g, x\big>$, where $g\sim N(0,I_n)$ is a standard Gaussian vector. 
The goal is to establish the following inequality:
For any set $T\subset \mathbb R^n$ and a point $y\in T$, we have
$$\frac{1}{3}[w(T)+\|y\|_2]\le \gamma(T) \le 2[w(T)+\|y\|_2].$$
I know the following equality may be helpful (follows from his proposition 7.5.2(e)):
$$w(T)=\frac{1}{2}w(T-T)=\frac{1}{2}\gamma(T-T).$$
So the problem boils down to finding the relations between $\gamma(T-T),\gamma(T)$, the points in $T$ that are farthest and nearest to the origin. But I don't know how to give a desired description (it's okay to replace the constants $\frac{1}{3}$ and $2$ with other constants).
Thanks in advance! 
 A: The lower bound:
$3E \sup_{x\in T}|\langle g,x\rangle|\ge E \sup_{x\in T}|\langle g,x\rangle|+2 \mathbb{E}|\langle g,y\rangle|\ge w(T)+2\sqrt{2/\pi}\|y\|\ge w(T)+\|y\|$
A: The upper bound:
Writing $x = x-y+y$ and using the triangle inequality, we have
$$E \sup_{x \in T} |\langle g, x \rangle | \le E \sup_{x \in T} | \langle g, x-y \rangle| + E| \langle g, y \rangle |.$$
The second term equals $\|y\|$. The first term is bounded by $\gamma(T-T) = 2 w(T)$.
Correction: the second term is $E|\langle g, y \rangle | = \|y\| E_{Z \sim N(0,1)}|Z| = \sqrt{2/\pi} \|y\| \le \|y\|$. Thanks to yxc for pointing this out.
A: This meant to be a comment, but I can't seem to do so.
Regarding angryavian's answer, the second term does NOT equal $\|y\|$. It is $\sqrt{\frac{2}{\pi}}\|y\|$, which is bounded above by $\|y\|$. For more information, one can refer to Eg. https://www.quora.com/If-Y-X-where-X-has-normal-distribution-N-0-1-what-is-the-density-function-expectation-and-variance-of-Y
One should also find $\mathbb{E}|\langle g,y\rangle|=\|y\|$ intuitively bothering: having this equality would mean that for $X=\langle g,y\rangle$, we have $ (\mathbb{E}|X|)^2=\mathbb{E}(X^2)$ and $$
\text{Var} |X|=\mathbb{E}(|X|^2)-(\mathbb{E}|X|)^2=\mathbb{E}(X^2)-\mathbb{E}(X^2)=0,
$$ which implies $|X|$ is constant with probability 1, and we know that this is not necessarily true.
