I don't know if I've made any non-trivial progress on this, and I'm not sure how to continue? Could someone please help? Thank you!

I just realized

https://artofproblemsolving.com/community/c6h1830479p12685616

Case 1 : $a,b,c,d > 0$ The inequality is equivalent to : $$ ab+ac+ad+bc+bd+cd \geq 2abcd + 4\sqrt{abcd} $$From the AMGM Inequality we get the following results : $$ 1 =(\frac{a^2 + b^2 +c^2 + d^2 -2}{2})^2 \geq (a^2+c^2-1)(b^2+d^2-1) \geq (2ac-1)(2bd-1) \Leftrightarrow ac + bd \geq 2abcd $$$$ ab+ad + bc + cd =(a+c)(b+d) \geq 4\sqrt{abcd} $$Combining these two inequality we get the desired result. Equality at $a=b=c=d=1$. Case 2 : Some of $a,b,c,d$ is zero, WLOG let $a \geq b \geq c \geq d=0$. Thus $abcd = 0$, then the inequality is equivalent to : $$ac + bc + ca \geq 0$$Which is true, and equality happens if and only if : $(b,c) =(0,0)$ Thus the other equality happens when : $$(a,b,c,d) =(2,0,0,0)\quad \text{and its permutations}$$

parmenides51 wrote: Let $a, b, c, d \ge 0$ such that $a^2 + b^2 + c^2 + d^2 = 4$. Prove that $$\frac{a + b + c + d}{2} \ge 1 + \sqrt{abcd}$$When does the equality hold? Leonard Giugiuc and Valmir B. Krasniqi Here we go !! so let us assume that $a+b=m$, $c+d=n$, $ab=x$ and $cd=y$. Now,$$\frac{a+b+c+d}{2}\geq 1+\sqrt{abcd}$$is eequivalent to$$\frac{m+n}{2}\geq 1+\sqrt{xy}.$$Let us define$$f(x,y)=xy-(\frac{m+ n-2}{2}).$$Now,$$\frac{\partial^2 f(x,y)}{\partial^2 x}=0$$and$$\frac{\partial^2 f(x,y)}{\partial^2 y}=0$$. Thus $f(x,y)$ is convex in variables $x$ and $y$. since, $x=ab$ and $y=bc$, therefore $0\leq x\leq \frac{m^2}{4}$ and $0\leq y\leq \frac{n^2}{4}.$ According to the proposition that an n-variable convex function, when considered as a single variable function, it attains the maximum values at its boundaries. so,$$\max f(x,y)=\max f(\alpha,\beta)$$where $\alpha \in \{0, \frac{m^2}{4}\}$ and $\beta \in \{0, \frac{n^2}{4}\}.$ Now we need to consider the following two cases. Case - 1 : when $x=\frac{m^2}{4}$ and $y=\frac{n^2}{4}$, we get, $a=b$ and $c=d$. Thus, we need to prove the following inequality, $$a+c=1+ac$$when $a^2+c^2=2$. on squaring both sides we get, $$a^2+c^2=1+a^2c^2$$or$$(a^2-1)^2\geq 0$$which is obvious. Case - 2 : when $x=0$ and $y=0$ or $abcd=0$. This forces one of the variables to be equal to zero. Let us assume that $d=0$. So we need to prove the following, $$a+b+c\ge 2$$when $a^2+b^2+c^2=4$. When we square both the sides we get, $$(a+b+c)^2=a^2+b^2+c^2+2\sum(ab)\ge 4=a^2+b^2+c^2$$or$$2\sum(ab)\ge 0$$, which is obvious. This ends our proof. $\blacksquare$

wait since $f$ is convex could we use jensens

vsamc wrote: wait since $f$ is convex could we use jensens $f$ is convex w.r.t $x=ab$ and $y=bc$....for us to apply Jensen, $f$ should be convex or concave w.r.t. $a,b,c$ or $d$

Note that $a^2+b^2+c^2+d^2 \ge 4\sqrt{abcd}$ implies $1\ge abcd\ge 0$. Next, squaring both sides, it suffices to prove \[ (a+b+c+d)^2 = 4 + 2(a+b)(c+d) + 2ab+2cd \ge 4 + 8\sqrt{abcd}+4abcd, \]which is equivalent to \[ (a+b)(c+d) +ab+cd \ge 4\sqrt{abcd}+2abcd. \]Now, using $ab+cd\ge 2\sqrt{abcd}$ by AM-GM, $(a+b)(c+d)\ge 2\sqrt{ab}\cdot 2\sqrt{cd}=4\sqrt{abcd}$, we find \[ (a+b)(c+d)+ab+cd\ge 6\sqrt{abcd}\ge 4\sqrt{abcd}+2abcd, \]since $\sqrt{abcd}\ge abcd$ as $1\ge abcd$.