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

Question

Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=abc$. Prove that: $$\frac{1}{7a+b}+\frac{1}{7b+c}+\frac{1}{7c+a}\leq\frac{\sqrt3}{8}$$

I tried C-S: $$\left(\sum_{cyc}\frac{1}{7a+b}\right)^2\leq\sum_{cyc}\frac{1}{(ka+mb+c)(7a+b)^2}\sum_{cyc}(ka+mb+c)=$$ $$=\sum_{cyc}\frac{(k+m+1)(a+b+c)}{(ka+mb+c)(7a+b)^2}.$$ Thus, it remains to prove that $$\sum_{cyc}\frac{k+m+1}{(ka+mb+c)(7a+b)^2}\leq\frac{3}{64abc},$$ but I did not find non-negative values of $k$ and $m$, for which the last inequality is true.

If we replace $7$ with $8$ so for $(a,b,c)||(28,1,5)$ this inequality would be wrong. Around this point the starting inequality is true, but we see that we can'not free use AM-GM because in AM-GM the equality occurs, when all variables are equal.

Thank you!

Answer 1

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

Answer 2

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.

Answer 3

Not a proof just a simplification for the proof wich already exists :

We prove the following inequality :

Let $a,b,c>0$ such that $abc=a+b+c$ and $a\geq b \geq c$ then we have : $$\sum_{cyc}\frac{1}{7a+b}\leq \sum_{cyc}\frac{1}{7b+a}$$

Proof :

We work with the following equivalent :

Let $a,b,c>0$ and $a\geq b \geq c$ then we have : $$\sqrt{\frac{abc}{a+b+c}}\sum_{cyc}\frac{1}{7a+b}\leq \sqrt{\frac{abc}{a+b+c}}\sum_{cyc}\frac{1}{7b+a}$$

Remains to show :

Let $a,b,c>0$ and $a\geq b \geq c$ then we have : $$\sum_{cyc}\frac{1}{7a+b}\leq \sum_{cyc}\frac{1}{7b+a}$$

We make the difference we get :

$$\sum_{cyc}\frac{1}{7a+b}-\sum_{cyc}\frac{1}{7b+a}=-\frac{(42 (a - b) (a - c) (b - c) (7 a^2 + 57 a b + 57 a c + 7 b^2 + 57 b c + 7 c^2))}{((7 a + b) (a + 7 b) (7 a + c) (a + 7 c) (7 b + c) (b + 7 c))}\leq 0$$

I think it simplify a part of the proof of River Li and maybe create a new approach to solve the initial inequality.

Answer 4

My second solution:

Remark: Using the Buffalo Way as well, but it is simpler.

Use @chenbai's substitutions $x=\frac{\sqrt{3}}{a}, y=\frac{\sqrt{3}}{b}, z=\frac{\sqrt{3}}{c}$. We have $xy+yz+zx=3$. We need to prove that $$\frac{xy}{x + 7y}+\frac{yz}{y + 7z}+\frac{zx}{z+7x}\leq\frac{3}{8}.$$

After homogenization, it suffices to prove that $$\Big(\frac{xy}{x + 7y}+\frac{yz}{y + 7z}+\frac{zx}{z+7x}\Big)^2 \leq \frac{9}{64}\frac{xy+yz+zx}{3}$$ or (clearing the denominators) $$f(x, y, z) \ge 0$$ where $f(x, y, z)$ is a homogeneous polynomial.

WLOG, assume $z = \min(x, y, z)$. Let $x = z + s, y = z + t$ for $s, t\ge 0$. We have $$f(z + s, z + t, z) = q_6z^6 + q_5 z^5 + q_4 z^4 + q_3 z^3 + q_2 z^2 + q_1 z + q_0$$ where \begin{align*} q_6 &= 81920 s^2-81920 s t+81920 t^2, \\ q_5 &= 106496 s^3-107520 s^2 t+279552 s t^2+106496 t^3, \\ q_4 &= 49664 s^4-155648 s^3 t+312832 s^2 t^2+489472 s t^3+49664 t^4, \\ q_3 &= 18816 s^5-97904 s^4 t-39200 s^3 t^2+650272 s^2 t^3+202480 s t^4+18816 t^5,\\ q_2 &= 14112 s^5 t-135143 s^4 t^2+239198 s^3 t^3+244873 s^2 t^4+42336 s t^5, \\ q_1 &= 2646 s^5 t^2-23590 s^4 t^3+91322 s^3 t^4+30870 s^2 t^5, \\ q_0 &= 147 s^5 t^3-1078 s^4 t^4+7203 s^3 t^5. \end{align*} It is easy to prove that $q_5, q_4, q_3, q_1, q_0 \ge 0$. It suffices to prove that $4q_3 q_1 \ge q_2^2$ or \begin{align} &s^2 t^3\Big(1002582336 s^7-9318167153 s^6 t+34882209252 s^5 t^2-77849923910 s^4 t^3\\ &\quad +108085842468 s^3 t^4+72267463855 s^2 t^5+11141602752 s t^6+531062784 t^7\Big) \ge 0 \end{align} which is true.

We are done.

Answer 5

This inequality is something hard

$ a\geq b\geq\sqrt{3}\geq c>0$ such that $abc=a+b+c$ and constraints $(C)$ and $(2)$ and $(3)$ then we have :

$$\frac{1}{7b+a}+\frac{1}{7a+c}+\frac{1}{7c+b}\leq \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\frac{1}{1+7\left(\frac{\frac{b}{a}+\frac{a}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3}{9}\right)^{\frac{1}{3}}}$$

And :

$$\frac{1}{7b+a}\leq \left(\frac{1}{b}\right)\frac{1}{1+7\left(\frac{\frac{b}{a}+\frac{a}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3}{9}\right)^{\frac{1}{3}}}\quad (C)$$

$$\frac{1}{7c+b}+\frac{1}{7a+c}\leq \left(\frac{1}{c}+\frac{1}{a}\right)\frac{1}{1+7\left(\frac{\frac{b}{a}+\frac{a}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3}{9}\right)^{\frac{1}{3}}}\quad (2)$$

Simplify the condition $(C)$ is easy comparing the function :

$$f(x)=\frac{1}{7x+1}$$

Or:

$$f\Big(\frac{6+\frac{a}{b}}{7}\Big)\leq f\left(\left(\frac{\frac{b}{a}+\frac{a}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3}{9}\right)^{\frac{1}{3}}\right)$$

For the constraint $(2)$ we know that :

$$f(a)=\frac{1}{7c+b}+\frac{1}{7a+c}-\left(\frac{1}{c}+\frac{1}{a}\right)\frac{1}{1+7\left(\frac{\frac{b}{a}+\frac{a}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3}{9}\right)^{\frac{1}{3}}}$$

is an increasing function (as the difference of a positive decreasing function and a negative increasing function) on $(\sqrt{3},\frac{\frac{3}{2}\sqrt{3}-\frac{1}{2}c+c}{c(\frac{3}{2}\sqrt{3}-\frac{1}{2}c)-1})$ with : $$b=\frac{a+c}{ac-1}$$

So we have to find an approximation of the roots !

We can choose $b\geq\frac{3}{2}\sqrt{3}-\frac{1}{2}c\quad (3)$

A bit of algebra and we get a polynomial with a root equal to $\sqrt{3}$ and one other root .Now the magic key is,the inequality and the constraints $(C),(2)$ are homogenous so we can introduce a coefficient and play with it keeping in mind the others constraints.

I continue later...Thanks!