Suppose $(a, b, c, d)$ is an extremum point of $E$.
If $a, b, c, d$ are not all zero, then by Fermat's theorem, $f(t)=E(a, b+t, c-t, d)$ must have $f^{\prime}(0)=0$. Note that
$$f^{\prime}(0)=\frac{-c-2 d}{2(c+d)^{\frac{3}{2}}}+\frac{1}{\sqrt{b+c}}-\frac{a}{2(a+b)^{\frac{3}{2}}}$$

Thus,
$$\frac{-c-2 d}{2(c+d)^{\frac{3}{2}}}+\frac{1}{\sqrt{b+c}}-\frac{a}{2(a+b)^{\frac{3}{2}}}=0$$

Similarly, for $g(t)=E(a+t, b, c, d-t)$, we have $g^{\prime}(0)=0$, i.e.,
$$\frac{-a-2 b}{2(a+b)^{\frac{3}{2}}}+\frac{1}{\sqrt{a+d}}-\frac{c}{2(c+d)^{\frac{3}{2}}}=0$$

Therefore, we have
$$\frac{1}{\sqrt{a+b}}+\frac{1}{\sqrt{c+d}}=\frac{1}{\sqrt{b+c}}+\frac{1}{\sqrt{a+d}}$$

By the monotonicity of $\frac{1}{\sqrt{x}}+\frac{1}{1-\sqrt{x}}$, we have $a+b=b+c$ or $a+b=a+d$, i.e., $a=c$ or $b=d$. By symmetry, assume $a=c$, the condition becomes $a+2 b+c=1$, at this point
$$\begin{aligned} E(a, b, a, d)= & \frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{a+b}}+\frac{a}{\sqrt{a+d}}+\frac{d}{\sqrt{d+a}}= \\ & \sqrt{a+b}+\sqrt{a+d} \leqslant \\ & \sqrt{2(a+b+a+d)}=\sqrt{2}<\frac{3}{2} \end{aligned}$$

If $a, b, c, d$ include 0, we assume $d=0$. Then
$$\begin{aligned} E(a, b, c)= & \frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\sqrt{c} \leqslant \\ & \frac{a}{\sqrt{a+b}}+\sqrt{(b+c)\left(\frac{b}{b+c}+1\right)}= \\ & \frac{a}{\sqrt{a+b}}+\sqrt{1-a+b} \leqslant \\ & \sqrt{(a+(1-a+b))\left(\frac{a}{a+b}+1\right)} \leqslant \\ & \sqrt{(1+b)(2-b)} \leqslant \\ & \frac{(1+b)+(2-b)}{2}=\frac{3}{2} \end{aligned}$$