Mixing Variables and $uvw$ help.
I'll post my solution, which I found eight years ago.
Let $f(a,b,c,d)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{12}{a+b+c+d}- 7$ and $a=\max\{a,b,c,d\}$.
Thus, $$f(a,b,c,d)-f\left(a,\sqrt[3]{bcd},\sqrt[3]{bcd},\sqrt[3]{bcd}\right)=$$
$$=\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{12\left(b+c+d-3\sqrt[3]{bcd}\right)}{(a+b+c+d)\left(a+3\sqrt[3]{bcd}\right)}\geq$$
$$\geq\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{12\left(b+c+d-3\sqrt[3]{bcd}\right)}{(\frac{b+c+d}{3}+b+c+d)\left(\frac{b+c+d}{3}+3\sqrt[3]{bcd}\right)}=$$
$$=\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{27\left(b+c+d-3\sqrt[3]{bcd}\right)}{(b+c+d)\left(b+c+d+9\sqrt[3]{bcd}\right)}.$$
We'll prove that $$\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{27\left(b+c+d-3\sqrt[3]{bcd}\right)}{(b+c+d)\left(b+c+d+9\sqrt[3]{bcd}\right)}\geq0.$$
Indeed, let $b+c+d=3u$, $bc+bd+cd=3v^2$ and $bcd=w^3$.
Thus, $$\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{27\left(b+c+d-3\sqrt[3]{bcd}\right)}{(b+c+d)\left(b+c+d+9\sqrt[3]{bcd}\right)}\geq0$$ it's $ g(v^2)\geq0,$
where $g$ is a linear increasing function.
Id est, $g$ gets a minimal value, when $v^2$ gets a minimal value,
which happens for equality case of two variables.
Since $g(v^2)\geq0$ is homogeneous inequality, it's enough to check one case only:
$c=d=1$, which after substitution $b=x^3$ gives
$$(x-1)^2(2x^7+x^6+18x^5-10x^4-50x^3+36x^2+26x+4)\geq0,$$ which is true.
Id est, $$f(a,b,c,d)\geq f\left(a,\sqrt[3]{bcd},\sqrt[3]{bcd},\sqrt[3]{bcd}\right)$$ and it's enough to prove that $f(a,b,b,b)\geq0$, where $a=\frac{1}{b^3}$, which gives
$$(b-1)^2(3b^6+6b^5+9b^4-9b^3-5b^2-b+3)\geq0,$$ which is true.
We can use also the Cassel's inequality:
Let $a$, $b$ and $w$ be sequences of $n$ positive numbers such that $1<m\leq\frac{a_k}{b_k}\leq M$ for any $k.$ Prove that:
$$\sum_{k=1}^nw_ka_k^2\sum_{k=1}^nw_kb_k^2\leq\frac{(M+m)^2}{4Mm}\left(\sum_{k=1}^nw_ka_kb_k\right)^2.$$
This inequality was here:
G.S. WATSON, Serial Correlation in Regression Analysis, Ph.D. Thesis, Dept. of Experimental
Statistics, North Carolina State College, Raleigh; Univ. of North Carolina, Mimograph Ser., No.
49, 1951, appendix 1.
In our case $w_k=2k-1$, $a_k=\sqrt{\frac{k+1}{k}}$, $b_k=\sqrt{\frac{k}{k+1}},$ $M=2$ and $m=1$, which gives:
$$\sum_{k=1}^n(2k-1)\frac{k+1}{k}\sum_{k=1}^n(2k-1)\frac{k}{k+1}\leq\frac{(2+1)^2}{4\cdot2\cdot1}\left(\sum_{k=1}^n(2k-1)\right)^2=\frac{9n^4}{8}.$$
Best Answer
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, our inequality is equivalent to
$\frac{3v^2\sum\limits_{cyc}(x^2+3v^2)^2}{(9uv^2-w^3)^2}\geq\frac{9}{4}$, which is $f(w^3)\geq0$, where $f$ is a concave function.
Thus, $f$ gets a minimal value for an extremal value of $w^3$, which happens in the following cases.
$z\rightarrow0^+$, $y=1$, which gives $(x-1)^2(4x^2+7x+4)\geq0$;
$y=z=1$, which gives $x(x-1)^2\geq0$. Done!