If $a+b+c=abc$ then $\sum\limits_{cyc}\frac{1}{7a+b}\leq\frac{\sqrt3}{8}$

let $x=\frac{\sqrt{3}}{a},y=\frac{\sqrt{3}}{b},z=\frac{\sqrt{3}}{c} \implies xy+yz+zx=3 $

with uvw method:

$3u=x+y+z,3v^2=xy+yz+zx,w^3=xyz\\ u\geqslant v \geqslant w \geqslant 0 , w^3 \leq 3uv^2 -2u^3+2\sqrt{(u^2-v^2)^3}, v=1 $

then inequality becomes:

$\frac{xy}{7y+x}+\frac{yz}{7z+y}+\frac{xz}{7x+z}\leq\frac{3}{8} \iff \\3(7y+x)(7z+y)(7x+z) \geqslant 8[xy(7z+y)(7x+z)+yz(7y+x)(7x+z)+xz(7y+x)(7z+y)]$ \

$LHS=3[7(7yz^2+xz^2+y^2z+7x^2z+7xy^2+x^2y)+344xyz]\\7yz^2+xz^2+y^2z+7x^2z+7xy^2+x^2y=4\sum(x^2y+y^2x)+3(yz^2+x^2z+xy^2-xz^2-y^2z-x^2y)=4\sum xy(x+y+z-z)+3(y-x)(z-x)(z-y)=4(x+y+z)\sum xy -4*3xyz+3(y-x)(z-x)(z-y) \\ LHS=3[7*4*3^2u+260w^3+3*7(y-x)(z-x)(z-y)]\\ RHS=8[\sum_{cyc}xy(7z+y)(7x+z)]=8[7\sum x^2y^2+57xyz(x+y+z)]\\= 8[ 7((xy+yz+xz)^2-2xyz(x+y+z))+57xyz(x+y+z)] \\=8(63+43*3u*w^3)$

then the inequality becomes:

$4(63u+65w^3-42-86uw^3) \geq 21(x-y)(z-x)(z-y) ....(3)$

now to prove

$63u+65w^3-42-86uw^3 \geq 0 \iff w^3 \leq \dfrac{63u-42}{86u-65}$

first to prove :

$\dfrac{63u-42}{86u-65} \geq \dfrac{2u-1}{4u-3} \iff (u-1)(80u-61) $it is true as $u\geq 1$

second to prove

$\dfrac{2u-1}{4u-3} \geq 3u-2u^3+2\sqrt{(u^2-1)^3} \iff \\ (2u^3-3u)(4u-3)+2u-1 \geq 2(4u-3)\sqrt{(u^2-1)^3} \iff (u-1)(8u^3+2u^2-10u+1) \geq 2(4u-3)(u^2-1)\sqrt{(u^2-1)} \iff (u-1)^2[(8u^3+2u^2-10u+1)^2-4(4u-3)^2(u+1)^2(u^2-1)] \geq 0 \iff (u-1)^2(32u^3-24u^2-44u+37) \geq 0 $

$(u-1)^2 \geq 0$ ,it remains $h(u)=32u^3-24u^2-44u+37>0 $ $h'(u)=96*u^2-48*u-44 ,$ let $h'(u)=0 $ we have $u_1=\dfrac{5\sqrt3+3}{12},u_2=-\dfrac{5\sqrt3-3}{12}<0$

it is easy to verify that $h_{min}=h(u_1)=\dfrac{25(9-5\sqrt3)}{9}>0$

so $63u+65w^3-42-86uw^3 \geq 0$ is true and when and only when $u=1$ it takes $0$. when $u=v \implies x=y=z$

check RHS of (3) , if $(x-y)(z-x)(z-y) \leq 0$ , then (3) is true. when $(x-y)(z-x)(z-y) \geq 0$ , square both sides, we need to porve :

$4^2(63u+65w^3-42-86uw^3)^2 \geq 21^2 [(x-y)(z-x)(z-y)]^2$...(4)

$[(x-y)(z-x)(z-y)]^2=27[4(u^2-v^2)^3-(w^3-3uv^2+2u^2)^2]=27[4(u^2-1)^3-(w^3-3u+2u^3)^2]$

then the (4) becomes: $A(u)w^6+B(u)w^3+C(u) \geq 0 \\ A(u)=43(2752u^2-4160u+1849) >0 as 4160^2-4*2752*1849=-3048192 <0 \\ B(u)=42(1134u^3-4128u^2+4171u-2080) \\ C(u)=441(63u^2-192u+172)>0 as 192^2-4*63*172=-6480 <0$

let $t=w^3\implies 0\leq t\leq 1, A(u)w^6+B(u)w^3+C(u)=At^2+Bt+C=f(t)$

when $-\dfrac{B}{2A} \leq 0 , f_{min}=f(0)=C(u)>0$ when $-\dfrac{B}{2A} \geq 1 ,f_{min}=f(1)=A(u)+B(u)+C(u)=(u-1)^2(47628u+67999) \geq 0 $

when $0 \leq -\dfrac{B}{2A} \leq 1 \iff 42(1134u^3-4128u^2+4171u-2080)<0 $ and $-42(1134u^3-4128u^2+4171u-2080) \leq 2*43*(2752*u^2-4160*u+1849) $

we will prove $B^2-4AC<0$

$-42(1134u^3-4128u^2+4171u-2080) \leq 2*43*(2752u^2-4160u+1849) \iff \\ 2(u-1)(23814u^2+55462u-35827) \geq 0$ it is always true

let $g(u)=1134u^3-4128u^2+4171u-2080$, it is trivial that

$1134u^3-4128u^2+4171u-1909.536 \geq g(u) \geq 1134u^3-4128u^2+4171*u-5979$

$1134*u^3-4128*u^2+4171*u-1909.536=((u-2.4)*(28350*u^2-35160*u+19891))/25$

$(28350*u^2-35160*u+19891)=0 $ no real root as $35160^2-4*28350*19891=-1019413800 <0$

$1134*u^3-4128*u^2+4171*u-5979=(u-3)*(1134*u^2-726*u+1993)$

it is trivial that $1134*u^2-726*u+1993$ no real root.

which means $g(u)$ only have one real root $u_3$ and $2.4 <u_3 <3$

$g'(u)=3402*u^2-8256*u+4171,8256/(2*3402)=1.213<1.3,g'(2)=1267>0 \implies g(u) $
is mono increase function when $u>2 \implies \\$

$B(u)<0 $ when $u<u_3 $

now we prove when $u<3, B^2-4AC \leq 0$

$B^2-4AC=190512(u-1)^2*(11907u^4-62874u^3+38688u^2+92442*u-86563)$

$11907u^4-62874u^3+38688u^2+92442u-86563=11907(u-3)(u^2-1)(u-1)-2(7623u^3-7437u^2-22407u+25421)\\7623*u^3-7437*u^2-22407*u+25421=(7623*u^3-7443*u^2-22410*u+25416)+(6u^2+3u+5) \\7623u^3-7443u^2-22410u+25416=9(847u^3-827u^2-2490u+2824)$

$(847u^3-827u^2-2490u+2824)'_u=2541u^2-1654u-2490=0 , $

we have two roots

$u_4=\dfrac{\sqrt{7011019}+827}{2541},\\u_5=\dfrac{-\sqrt{7011019}+827}{2541} <0 $

it is easy to verify that $(847u^3-827u^2-2490u+2824)_{min}=847u_4^3-827u_4^2-2490u_4+2824>38>0$

so $B^2-4AC \leq 0$ is true .

QED

The Buffalo Way works.

After homogenization, it suffices to prove that $f(a,b,c)\ge 0$ where $f(a,b,c)$ is a polynomial given by \begin{align} f(a,b,c) &= 64abc(7a+b)^2(7b+c)^2(7c+a)^2\\ &\quad \times\left(\frac{3}{64}\frac{a+b+c}{abc} - \left(\frac{1}{7a+b} + \frac{1}{7b+c} + \frac{1}{7c+a}\right)^2\right). \end{align}

WLOG, assume that $c = \min(a, b, c).$ There are two possible cases:

1) $c \le b\le a$: Let $c = 1, \ b = 1+s, \ a = 1+s+t; \ s,t\ge 0$. $f(1+s+t, 1+s, 1)$ is a polynomial in $s, t$ with non-negative coefficients. So $f(1+s+t, 1+s, 1)\ge 0.$

2) $c \le a\le b$: Let $c =1, \ a=1+s, \ b=1+s+t; \ s,t\ge 0$. We have \begin{align} f(1+s, 1+s+t, 1) = a_5t^5 + a_4t^4 + a_3t^3 + a_2t^2 + a_1t + a_0 \end{align} where \begin{align} a_5 &= 147\, s^2 - 784\, s + 6272,\\ a_4 &= 2940\, s^3 - 16583\, s^2 + 53648\, s + 82432 ,\\ a_3 &= 19551\, s^4 - 94494\, s^3 - 65760\, s^2 + 185344\, s + 139264,\\ a_2 &= 49686\, s^5 - 68407\, s^4 - 242656\, s^3 + 13824\, s^2 + 220160\, s + 81920,\\ a_1 &= 51744\, s^6 + 97584\, s^5 + 88848\, s^4 + 173056\, s^3 + 211968\, s^2 + 81920\, s ,\\ a_0 &= 81920\, s^2 + 270336\, s^3 + 344576\, s^4 + 224640\, s^5 + 87296\, s^6 + 18816\, s^7. \end{align} It is easy to obtain that $a_5, a_4, a_1, a_0 \ge 0$. Thus, we have $$f(1+s, 1+s+t, 1)\ge (2\sqrt{a_5a_1} + a_3)t^3 + (2\sqrt{a_4a_0} + a_2)t^2.$$

It suffices to prove that $2\sqrt{a_5a_1} + a_3 \ge 0$ and $2\sqrt{a_4a_0} + a_2 \ge 0$.

Note that \begin{align} 2\sqrt{a_5a_1} + a_3 &= \Big(2\sqrt{a_5a_1} - \frac{1}{3}\cdot 94494\, s^3 - \frac{1}{3} \cdot 65760\, s^2\Big)\\ &\quad + \Big(a_3 + \frac{1}{3}\cdot 94494\, s^3 + \frac{1}{3} \cdot 65760\, s^2\Big) \end{align} and \begin{align} 2\sqrt{a_4a_0} + a_2 &= \Big(2\sqrt{a_4a_0} - \frac{1}{2}\cdot 68407\, s^4 - \frac{1}{2}242656\, s^3\Big)\\ &\quad + \Big(a_2 + \frac{1}{2}\cdot 68407\, s^4 + \frac{1}{2}242656\, s^3\Big). \end{align} It suffices to prove that \begin{align} 2\sqrt{a_5a_1} - \frac{1}{3}\cdot 94494\, s^3 - \frac{1}{3} \cdot 65760\, s^2 &\ge 0,\\ a_3 + \frac{1}{3}\cdot 94494\, s^3 + \frac{1}{3} \cdot 65760\, s^2&\ge 0,\\ 2\sqrt{a_4a_0} - \frac{1}{2}\cdot 68407\, s^4 - \frac{1}{2}242656\, s^3 &\ge 0,\\ a_2 + \frac{1}{2}\cdot 68407\, s^4 + \frac{1}{2}242656\, s^3 &\ge 0. \end{align} All of them can be reduced to polynomial inequalities in $s$ and not hard to prove. This completes the proof.