14
$\begingroup$

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!

$\endgroup$
  • 6
    $\begingroup$ FOr those fans who hate abbreviations, C-S is Cauchy-Schwartz $\endgroup$ – Mark Fischler Mar 21 '17 at 20:10
  • 5
    $\begingroup$ 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. $\endgroup$ – Mark Fischler Mar 21 '17 at 22:57
  • 1
    $\begingroup$ @W-t-P It's cyclic and not symmetric. $\endgroup$ – Michael Rozenberg Mar 24 at 20:42
  • $\begingroup$ @MichaelRozenberg Maybe BW helps? Though it doesn't seem to be particularly elegant... $\endgroup$ – user574848 Apr 1 at 10:18
  • 1
    $\begingroup$ @user574848 I tried. I think BW does not help here. $\endgroup$ – Michael Rozenberg Apr 1 at 20:57
1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ That's brute. Though it is legitimate as a proof, I think the OP needs a more elegant method. $\endgroup$ – Trebor Jun 3 at 8:59
  • $\begingroup$ How long did this take you? :D $\endgroup$ – Maximilian Janisch Jun 3 at 10:18
  • $\begingroup$ @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. $\endgroup$ – Michael Rozenberg Jun 5 at 6:10
  • $\begingroup$ 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.:) $\endgroup$ – chenbai Jun 6 at 5:58
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.