The condition gives $$1=\frac{abc(a+b+c)}{ab+ac+bc}$$
Id est, we need to prove that
$$ab+ac+bc\geq\frac{3abc(a+b+c)}{ab+ac+bc}$$ or
$$(ab+ac+bc)^2\geq3abc(a+b+c)$$ or
$$\sum_{cyc}(a^2b^2+2a^2bc)\geq3\sum_{cyc}a^2bc$$ or
$$\sum_{cyc}c^2(a-b)^2\geq0.$$
If you want to prove that $abc\geq1$ then it's impossible because it's wrong.
Indeed, $$abc\geq1$$ it's
$$abc\geq\left(\sqrt{\frac{abc(a+b+c)}{ab+ac+bc}}\right)^3$$ or
$$ab+ac+bc\geq(a+b+c)\sqrt[3]{abc},$$ which is wrong for $a\rightarrow+\infty.$
We can use the Vasc's RCF Theorem. It's like Jensen, but it's not Jensen.
Also, since $f(x)=\frac{1}{(1+e^x)^2}$ has an unique inflection point, we can use Jensen with Karamata, but it's not so nice solution. I am ready to show, if you want.
Indeed, $$f''(x)=\frac{4e^x\left(e^x-\frac{1}{2}\right)}{(1+e^x)^4}.$$
Thus, $f$ is a convex function on $[-\ln2,+\infty)$ and a concave function on $(-\infty,-\ln2]$.
We need to prove that $$\sum_{cyc}f(x)\geq0,$$ where $x+y+z+t=0$.
Now, let $x\geq y\geq z\geq t.$
We'll consider the following cases.
- $x\geq y\geq z\geq t\geq-\ln2.$
Thus, by Jensen $$\sum_{cyc}f(x)\geq4f\left(\frac{x+y+z+t}{4}\right)=4f(0)=1.$$
- $x\geq y\geq z\geq-\ln2\geq t$.
Thus, by Jensen again:
$$\sum_{cyc}f(x)\geq3f\left(\frac{x+y+z}{3}\right)+f(t)=3f\left(\frac{-t}{3}\right)+f(t).$$
Thus, it's enough to prove that
$$3f\left(\frac{-t}{3}\right)+f(t)\geq0,$$ which is $$\sum_{cyc}\frac{1}{(1+a)^2}\geq1,$$ where $b=c=a$ and $d=\frac{1}{a^3}$ or
$$\frac{3}{(1+a)^2}+\frac{1}{\left(1+\frac{1}{a^3}\right)^2}\geq1$$ or
$$(a-1)^2(3a^2-2a+2)\geq0,$$ which is obvious.
- $x\geq y\geq-\ln2\geq z\geq t$.
Thus, by Jensen again we have:
$$f(x)+f(y)\geq2f\left(\frac{x+y}{2}\right)^2=\frac{2}{\left(1+e^{\frac{x+y}{2}}\right)^2}=\frac{2}{(1+\sqrt{ab})^2}.$$
Also, since $$(-\ln2,\ln2+z+t)\succ(z,t),$$ by Karamata we obtain:
$$f(z)+f(t)\geq f(-\ln2)+f(\ln2+z+t)=$$
$$=\frac{1}{\left(1+e^{-\ln2}\right)^2}+\frac{1}{\left(1+e^{\ln+z+t}\right)^2}=\frac{4}{9}+\frac{1}{(1+2cd)^2}.$$
Let $\sqrt{ab}=u$.
Thus, it's enough to prove in this case that
$$\frac{4}{9}+\frac{1}{\left(1+\frac{2}{u^2}\right)^2}+\frac{2}{(1+u)^2}\geq1$$ and since $$\left(\frac{1}{\left(1+\frac{2}{u^2}\right)^2}+\frac{2}{(1+u)^2}\right)'=\frac{4(u^3-2)(u^3+6u^2+4)}{(u+1)^3(u^2+2)^3},$$ it's enough to prove the last inequality for $u=\sqrt[3]2,$ which gives
$$\frac{4}{9}+\frac{1}{\left(1+\frac{2}{u^2}\right)^2}+\frac{2}{(1+u)^2}=\frac{4}{9}+\frac{1}{(1+\sqrt[3]2)^2}+\frac{2}{(1+\sqrt[3]2)^2}>\frac{4}{9}+\frac{3}{(1+1.3)^2}>1.$$
- $x\geq-\ln2\geq y\geq z\geq t$.
Thus, since $$\left(-\ln2,-\ln2,2\ln2+y+z+t\right)\succ(y,z,t),$$ by Karamata again we obtain:
$$f(y)+f(z)+f(t)\geq2f(-\ln2)+f(2\ln2+y+z+t)=$$
$$=\frac{8}{9}+\frac{1}{\left(1+e^{2\ln2+y+z+t}\right)^2}=\frac{8}{9}+\frac{1}{(1+4bcd)^2}.$$
Id est, it's enough to prove that:
$$\frac{1}{(1+a)^2}+\frac{8}{9}+\frac{1}{\left(1+\frac{4}{a}\right)^2}\geq1$$ or
$$8a^4+8a^3-15a^2+32a+128\geq0,$$ which is obvious.
The case $-\ln2\geq x\geq y\geq z\geq t$ is impossible and we are done!
Best Answer
This proof uses the rearrangement inequality in addition to AM-GM. After multiplying by $abcd$ our problem is equivalent to solving $$ a^2cd + ab^2d + abc^2 + bcd^2 \leq 4$$ Let $\{a,b,c,d\}=\{w,x,y,z\}$ with $w \ge x \ge y \ge z$. We have $$ a^2cd + ab^2d + abc^2 + bcd^2 = a(acd)+b(abd)+c(abc)+d(bcd)$$ and by the rearrangement equality $$ a(acd)+b(abd)+c(abc)+d(bcd) \le w(wxy)+x(wxz)+y(wyz)+z(xyz)$$ $$ = (wx+yz)(wy+xz)$$ By AM-GM, we get $$ (wx+yz)(wy+xz) \le \left(\frac{wx+yz+wy+xz}{2}\right)^2 = \frac{1}{4} \left((w+z)(x+y)\right)^2$$ Another AM-GM gives us $$ \frac{1}{4} ((w+z)(x+y))^2 \le \frac{1}{4} \left(\left(\frac{w+x+y+z}{2}\right)^2\right)^2 = \frac{1}{4} \left(\left(\frac{4}{2}\right)^2\right)^2 = 4$$ Thus we have $$ a^2cd + ab^2d + abc^2 + bcd^2 \leq 4$$ and furthermore, we have $$ \frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \le \frac{4}{abcd}$$ You might be able to tweak this to avoid the use of rearrangement, but I couldn't find a way.