<s>[INST] 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.
 [/INST] Let $a+b+c+d>4$, $a=kx$, $b=ky$, $c=kz$ and $d=kt$ such that $k>0$ and $$x+y+z+t=4.$$
Thus, $$k(x+y+z+t)>4,$$ which gives $$k>1.$$
But, $$5=a^2+b^2+c^2+d^2 + abcd=k^2(x^2+y^2+z^2+t^2)+k^4xyzt>x^2+y^2+z^2+t^2+xyzt,$$
which is a contradiction because we'll prove now that $$x^2+y^2+z^2+t^2+xyzt\geq5$$ or
$$16(x^2+y^2+z^2+t^2)(x+y+z+t)^2+256xyzt\geq5(x+y+z+t)^4.$$
Indeed, let $x=\min\{x,y,z,t\}$, $y=x+u$, $z=x+v$ and $t=x+w$.
Thus, $$16(x^2+y^2+z^2+t^2)(x+y+z+t)^2+256xyzt-5(x+y+z+t)^4=$$
$$=32\sum_{cyc}(3u^2-2uv)x^2+16\sum_{cyc}(5u^3-u^2v-u^2w)x+$$
$$+(u+v+w)^2\sum_{cyc}(11u^2-10v^2)\geq0$$ and we are done! 
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