[proofplan] The proof has two independent parts. First, the coordinatewise inequality $a_i\le b_i$ makes every defining summand for $E(b_1,\dots,b_n)$ no larger than the corresponding summand for $E(a_1,\dots,a_n)$, so the set inclusion is literal; it is symplectic because it is the restriction of the identity map on $(\mathbb C^n,\omega_0)$. Second, the ellipsoid contains a standard ball whose capacity is the smallest axis parameter $\min_i a_i$. Composing this ball inclusion, or any other symplectic ball embedding into the ellipsoid, with a proposed embedding into the cylinder gives a ball-cylinder symplectic embedding, and Gromov non-squeezing forces the stated radius bound. [/proofplan]

[step:Compare the defining inequalities coordinate by coordinate] Let \begin{align*} \iota:E(a_1,\dots,a_n)\to E(b_1,\dots,b_n) \end{align*} denote the ordinary inclusion map, provided it is well-defined. Fix $z=(z_1,\dots,z_n)\in E(a_1,\dots,a_n)$. Since $a_i\le b_i$ and $a_i,b_i>0$, we have \begin{align*} \frac{\pi |z_i|^2}{b_i}\le \frac{\pi |z_i|^2}{a_i} \end{align*} for every $i\in\{1,\dots,n\}$. Summing these inequalities gives \begin{align*} \sum_{i=1}^n\frac{\pi |z_i|^2}{b_i}\le \sum_{i=1}^n\frac{\pi |z_i|^2}{a_i}<1. \end{align*} Therefore $z\in E(b_1,\dots,b_n)$, and hence \begin{align*} E(a_1,\dots,a_n)\subset E(b_1,\dots,b_n). \end{align*} The map $\iota$ is the restriction of the identity map \begin{align*} \operatorname{id}_{\mathbb C^n}:\mathbb C^n\to\mathbb C^n. \end{align*} Since $\operatorname{id}_{\mathbb C^n}^*\omega_0=\omega_0$, the restricted map satisfies \begin{align*} \iota^*(\omega_0|_{E(b_1,\dots,b_n)})=\omega_0|_{E(a_1,\dots,a_n)}. \end{align*} Thus $\iota$ is a symplectic embedding. [/step]

[step:Find the largest coordinate-free standard ball contained in the ellipsoid]Define \begin{align*} m:=\min_{1\le i\le n}a_i. \end{align*} Since each $a_i>0$, we have $m>0$. Let \begin{align*} \rho:=\sqrt{\frac{m}{\pi}}. \end{align*} We claim that \begin{align*} B^{2n}(\rho)\subset E(a_1,\dots,a_n). \end{align*} Indeed, if $z=(z_1,\dots,z_n)\in B^{2n}(\rho)$, then \begin{align*} \sum_{i=1}^n |z_i|^2=|z|^2<\rho^2=\frac{m}{\pi}. \end{align*} Because $m\le a_i$ for every $i$, we have \begin{align*} \frac{\pi |z_i|^2}{a_i}\le \frac{\pi |z_i|^2}{m} \end{align*} for every $i\in\{1,\dots,n\}$. Summing gives \begin{align*} \sum_{i=1}^n\frac{\pi |z_i|^2}{a_i}\le \frac{\pi}{m}\sum_{i=1}^n |z_i|^2<1. \end{align*} Thus $z\in E(a_1,\dots,a_n)$, proving the claimed ball inclusion.[/step]

[guided]The goal of this step is to identify a standard ball that is definitely contained in the ellipsoid without choosing a preferred coordinate axis. Define \begin{align*} m:=\min_{1\le i\le n}a_i. \end{align*} This number is positive because all $a_i$ are positive. The natural ball radius is \begin{align*} \rho:=\sqrt{\frac{m}{\pi}}, \end{align*} because the standard ball $B^{2n}(\rho)$ has capacity $\pi\rho^2=m$. We now verify the containment directly from the definitions. Take \begin{align*} z=(z_1,\dots,z_n)\in B^{2n}(\rho). \end{align*} By definition of the Euclidean ball in $\mathbb C^n\cong\mathbb R^{2n}$, \begin{align*} \sum_{i=1}^n |z_i|^2=|z|^2<\rho^2=\frac{m}{\pi}. \end{align*} Since $m\le a_i$ for each index $i$, division by positive numbers gives \begin{align*} \frac{1}{a_i}\le \frac{1}{m}. \end{align*} Multiplying by the nonnegative quantity $\pi |z_i|^2$ gives \begin{align*} \frac{\pi |z_i|^2}{a_i}\le \frac{\pi |z_i|^2}{m}. \end{align*} After summing over all $i\in\{1,\dots,n\}$, we obtain \begin{align*} \sum_{i=1}^n\frac{\pi |z_i|^2}{a_i}\le \frac{\pi}{m}\sum_{i=1}^n |z_i|^2<1. \end{align*} This is exactly the defining inequality for $z\in E(a_1,\dots,a_n)$. Hence \begin{align*} B^{2n}\left(\sqrt{\frac{m}{\pi}}\right)\subset E(a_1,\dots,a_n). \end{align*}[/guided]

[step:Compose the contained ball with the assumed cylinder embedding] Assume that \begin{align*} \Phi:E(a_1,\dots,a_n)\to Z^{2n}(R) \end{align*} is a symplectic embedding. Let \begin{align*} j:B^{2n}(\rho)\to E(a_1,\dots,a_n) \end{align*} denote the ordinary inclusion map from the previous step. As before, $j$ is the restriction of the identity map on $\mathbb C^n$, so $j$ is symplectic. Therefore the composition \begin{align*} \Phi\circ j:B^{2n}(\rho)\to Z^{2n}(R) \end{align*} is a symplectic embedding. [/step]

[step:Apply non-squeezing to the composed ball-cylinder embedding] The hypotheses of [citetheorem:10075] apply to the symplectic embedding \begin{align*} \Phi\circ j:B^{2n}(\rho)\to Z^{2n}(R), \end{align*} because $n\ge 2$, $\rho>0$, $R>0$, and both source and target carry the standard symplectic form $\omega_0$. Hence \begin{align*} \rho\le R. \end{align*} Multiplying by $\pi$ after squaring both sides gives \begin{align*} \pi\rho^2\le \pi R^2. \end{align*} Since $\rho^2=m/\pi$, this becomes \begin{align*} m\le \pi R^2. \end{align*} By the definition of $m$, this is \begin{align*} \min_{1\le i\le n}a_i\le \pi R^2. \end{align*} [/step]

[step:Repeat the same composition argument for any embedded ball in the ellipsoid] Let $r>0$, and suppose \begin{align*} \psi:B^{2n}(r)\to E(a_1,\dots,a_n) \end{align*} is any symplectic embedding. With the symplectic embedding $\Phi:E(a_1,\dots,a_n)\to Z^{2n}(R)$ already fixed, \begin{align*} \Phi\circ\psi:B^{2n}(r)\to Z^{2n}(R) \end{align*} is a symplectic embedding. Again applying [citetheorem:10075], with the same verification that $n\ge 2$, $r>0$, $R>0$, and the standard symplectic form is used on both Euclidean domains, gives \begin{align*} r\le R. \end{align*} Equivalently, \begin{align*} \pi r^2\le \pi R^2. \end{align*} Thus every symplectically embedded standard ball inside the ellipsoid obeys the non-squeezing radius bound imposed by the cylinder. This completes the proof. [/step]