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

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!

• FOr those fans who hate abbreviations, C-S is Cauchy-Schwartz – Mark Fischler Mar 21 '17 at 20:10
• As a good start, take $a=\sqrt{3}x, b = \sqrt{3}y, c = \sqrt{3}z$ so that you are working with $x+y+z = 3xyz$ and the extremum will occur at $(1,1,1)$. Makes the horrendous algebra a little cleaner. – Mark Fischler Mar 21 '17 at 22:57
• @W-t-P It's cyclic and not symmetric. – Michael Rozenberg Mar 24 at 20:42
• @MichaelRozenberg Maybe BW helps? Though it doesn't seem to be particularly elegant... – user574848 Apr 1 at 10:18
• @user574848 I tried. I think BW does not help here. – Michael Rozenberg Apr 1 at 20:57

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.

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

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

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

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

• That's brute. Though it is legitimate as a proof, I think the OP needs a more elegant method. – Trebor Jun 3 at 8:59
• How long did this take you? :D – Maximilian Janisch Jun 3 at 10:18
• @chenbai Thank you for your trying. I also have a proof because this inequality it' just a cubic inequality of $v^2$, which gives a very ugly solution. I looked for a proof, which we can write during a competition. – Michael Rozenberg Jun 5 at 6:10
• yes, it is a ugly proof. I knew it is impossible to finish in the competition. and I do have same hope to see some elegant method. Hope this ugly proof may take some better ones come. it takes me a longtime indeed.:) – chenbai Jun 6 at 5:58

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.