Inspired by Lakshya Sinha

Using Holder's, we have that

$\biggl(a^{2016} + b^{2016} + c^{2016} + d^{2016} \biggr)^{\frac{1}{2016}} \biggl(a^{2016} + b^{2016} + c^{2016} + d^{2016} \biggr)^{\frac{1}{2016}} \underbrace{\biggl(1+1+1+1\biggr)^{\frac{1}{2016}} \cdots \biggl(1+1+1+1\biggr)^{\frac{1}{2016}}}_{\text{2014 products}} \geq a^2 +b^2 +c^2 + d^2$

The sums in the first two parentheses equal 4032, while the remaining 2014 sums equal 4:

$\biggl(4032\biggr)^{\frac{2}{2016}} \biggl( 4\biggr)^{\frac{2014}{2016}} \geq a^2 +b^2 +c^2 + d^2$

$4.027 \geq a^2 +b^2 +c^2 + d^2$

The Jensen's inequality states that

$\frac{f(a_{1})+f(a_{2})+ \cdots + f(a_{n})}{n} \geq f(\frac{a_{1}+a_{2}+ \cdots a_{n}}{n})$ if $f(x)$ is convex.

Let $f(x)=x^{1008}$ . Note that $f'(x)=1008x^{1007}, f''(x)=1008.1007.x^{1006} > 0$ since $x$ is positive real. So $f(x)$ is convex.

Therefore,

$f(\frac{a^{2}+b^{2}+c^{2}+d^{2}}{4}) \leq \frac{f(a^{2})+f(b^{2})+f(c^{2})+f(d^{2})}{4}$

$({\frac{a^{2}+b^{2}+c^{2}+d^{2}}{4}})^{1008} \leq (\frac{a^{2016}+b^{2016}+c^{2016}+d^{2016}}{4})$

$({\frac{a^{2}+b^{2}+c^{2}+d^{2}}{4}})^{1008} \leq (\frac{4032}{4})$

$a^{2}+b^{2}+c^{2}+d^{2} \leq 4.1008^{\frac{1}{1008}} = 4.0275 \approx \boxed{4.028}$

Equality holds when $a=b=c=d=1008^{\frac{1}{2016}}$

The simplest solution is the power mean. The expression 4(1008)^(1/1008) is obtained in an instant.