Proving $abcd+3\geq a+b+c+d$ 
If $a,b,c,d$ are non negatives and $a^2+b^2+c^2+d^2=3$ prove that  $$abcd+3\ge a+b+c+d$$

The inequality is not as simple as it looks.The interesting part is that the equality occurs when $a=0,b=c=d=1$ upto permutation .(I don't know if there are further equality cases.)
I tried rewriting the inequality as $$a^2+a(bcd-1)+b^2+c^2+d^2-b-c-d\ge 0$$ As its a quadratic in $a$ it suffices to show  the discriminant$$\Delta_a={(bcd-1)}^2-4(b^2+c^2+d^2-b-c-d)\le 0$$ which unfortunately is wrong (when $a=\sqrt{3},b=c=d=0$)
P.S ;I am not aware of using Lagranges multipliers
Its from here
Update :An answer has been posted in the link above (AOPS) similar to Dr Mathva's answer

Due to symmetry, assume wlog that $a\geqslant b\geqslant c\geqslant d\geqslant 0$. Since I will use this fact later, I am proving it before the cases:

Lemma: We have that $2\geqslant a+d$.

In fact, this follows from Cauchy-Schwarz:  $$2= 2\sqrt{\frac{a^2+b^2+c^2+d^2}{3}}\geqslant 2\sqrt{\frac{a^2+3d^2}{3}}=\sqrt{1+\frac13}\cdot \sqrt{a^2+3d^2}\geqslant a+d$$
We will now consider some cases depending on the position of the number $1$:

*

*Case 1: $1\geqslant a\geqslant b\geqslant c\geqslant d\geqslant 0$. Observe that one may rewrite the inequality as $$3+abcd-(a+b+c+d)=(1-a)(1-b)+(1-c)(1-d)+(1-ab)(1-cd) $$ Which is clearly positive.


*Case 2: $a\geqslant 1\geqslant b\geqslant c\geqslant d\geqslant 0$. If $ab\geqslant 1$, we might proceed as in the next case. So we will work with $ab<1$. Thus \begin{align*}3+abcd-(a+b+c+d)&=(1-a)(1-b)+(1-c)(1-d)+(1-ab)(1-cd)\\&\geqslant \underbrace{(1-a)}_{<0}(1-b)+(1-c)(1-d)\\&\geqslant (1-a)(1-c)+(1-c)(1-d)\\&=(1-c)(2-(a+d))\geqslant 0  \end{align*} Where the last inequality follows from the lemma.


*Case 3: $a\geqslant b\geqslant 1\geqslant c\geqslant d\geqslant 0$. This implies that \begin{align*}6+2abcd-2(a+b+c+d)&=a^2+b^2+c^2+d^2+3+2abcd-2(a+b+c+d)\\
&=(a-1)^2+(b-1)^2+(c+d-1)^2+2cd(ab-1)\geqslant 0\end{align*} Or, equivalently $3+abcd\geqslant a+b+c+d$.


*Case 4: $a\geqslant b\geqslant c\geqslant 1\geqslant d\geqslant 0$. As @dezdichado noticed, this case is straightforward, since it forces directly $a=b=c=1, d=0$ due to the constraint $a^2+b^2+c^2+d^2=3$.


*Case 5: $a\geqslant b\geqslant c\geqslant d\geqslant 1$. Clearly impossible, since this would yield $a^2+b^2+c^2+d^2\geqslant 4>3$.
Done!
