For positive a,b,c,d, if $a^2+b^2+c^2+d^2+abcd=5$, show $a+b+c+d\leq 4$. One can use Lagrange multiplier, but I am looking for a more elementary proof.
I try to find the maximum of $a+b+c+d$ and follow the standard approach.
Construct $a+b+c+d+\lambda (a^2+b^2+c^2+d^2+abcd-5)$. 
One can obtain $1+2a\lambda +bcd =0$ and so on, and eventually have $a=b=c=d$. 
Then you are done.

Alternative solution:
Fact 1: For $a, b, c, d \ge 0$, it holds that
$$a^2 + b^2 + c^2 + d^2 + abcd - 5 - 3(a+b+c+d - 4) \ge 0.\tag{1}$$
From Fact 1, the desired result follows.
Edit 2022/02/14: A nice proof of Fact 1 is given by mudok@AoPS.
See: #4 in https://artofproblemsolving.com/community/c6t243f6h2780412_hard_inequality

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Proof of Fact 1:
From Vasc's Equal Variable Theorem [1, Corollary 1.9], we only need to prove the case
when $a = b = c$.
It suffices to prove that
$d^2+(a^3-3)d+3a^2-9a+7 \ge 0$.
Since $3a^2 - 9a + 7 > 0$, by AM-GM, we have
$d^2+(a^3-3)d+3a^2-9a+7 \ge (a^3-3)d + 2d\sqrt{3a^2-9a+7}$.
It suffices to prove that $a^3 - 3 + 2\sqrt{3a^2-9a+7} \ge 0$.
If $a > \frac{3}{2}$, then $a^3 > 3$ and the inequality is true.
If $a \le \frac{3}{2}$, then it suffices to prove that
$4(3a^2-9a+7) - (a^3 - 3)^2 \ge 0$ or $(-a^4 - 2a^3 - 3a^2 + 2a + 19)(a-1)^2 \ge 0$.
We have $-a^4 - 2a^3 - 3a^2 + 2a + 19 \ge - (3/2)^4 - 2\cdot (3/2)^3 - 3\cdot (3/2)^2 + 19 > 0$.
We are done.
Reference
[1] Vasile Cirtoaje, “The Equal Variable Method”, J. Inequal. Pure and Appl. Math., 8(1), 2007.
https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf
