Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that: $$\frac{a}{\sqrt[4]{8(b^4+c^4)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}$$

Nesbitt's inequality is the following:

Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that: $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2},$$ which follows from C-S: $$\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{a^2}{ab+ac}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(ab+ac)}=\frac{(a+b+c)^2}{2(ab+ac+bc)}\geq\frac{3}{2},$$ but this way does not help for the starting inequality.

There is a nice solution for the following inequality, which was in our test six months ago.

Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\frac{a}{\sqrt[3]{4(b^3+c^3)}}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$$

By C-S $$\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{(b+c)^2}{ab+ac+2bc}.$$ Thus, it's enough to prove that $$\frac{a}{\sqrt[3]{4(b^3+c^3)}}+\frac{(b+c)^2}{ab+ac+2bc}\geq\frac{3}{2}$$ or $$2(b+c)a^2-(3(b+c)\sqrt[3]{4(b^3+c^3)}-4bc)a+2(b^2-bc+c^2)\sqrt[3]{4(b^3+c^3)}\geq0.$$ Thus, it's enough to prove that $$16(b+c)(b^2-bc+c^2)\sqrt[3]{4(b^3+c^3)}\geq\left(3(b+c)\sqrt[3]{4(b^3+c^3)}-4bc\right)^2$$ or $$16\sqrt[3]{4(b^3+c^3)^4}\geq\left(3(b+c)\sqrt[3]{4(b^3+c^3)}-4bc\right)^2$$ and since $$3(b+c)\sqrt[3]{4(b^3+c^3)}>4bc,$$ it remains to prove that $$4\sqrt[3]{2(b^3+c^3)^2}\geq3(b+c)\sqrt[3]{4(b^3+c^3)}-4bc$$ and after using $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz),$$ where $$x^2+y^2+z^2-xy-xz-yz\neq0$$ we need to prove that $$128(b^3+c^3)^2-108(b+c)^3(b^3+c^3)+64b^3c^3+288(b+c)(b^3+c^3)bc\geq0.$$ Now, let $b^2+c^2=2kbc$. Hence, we need to prove that: $$128(2k+2)(2k-1)^2-108(2k+2)^2(2k-1)+64+288(2k+2)(2k-1)\geq0$$ or $$(k-1)^2(10k+11)\geq0.$$ Done!

But this way gives a wrong inequality again.

  • $\begingroup$ from where does it come? $\endgroup$ – Dr. Sonnhard Graubner May 15 '17 at 13:15
  • $\begingroup$ @Dr. Sonnhard Graubner It's my problem. $\endgroup$ – Michael Rozenberg May 15 '17 at 13:29
  • $\begingroup$ oh nice i will try it $\endgroup$ – Dr. Sonnhard Graubner May 15 '17 at 13:30
  • $\begingroup$ Perhaps the following may be true, but this is just a mere speculation: $$\forall n \in \mathbb{Z^+}, \ \frac{a}{\big(2^{n - 1}(b^n + c^n)\big)^{1/n}} + \frac{b}{a + c} + \frac{c}{a + b} \geqslant \frac{3}{2}$$ $\endgroup$ – Feeds Jan 15 '18 at 2:09
  • $\begingroup$ For $n=5$ it's wrong already. $\endgroup$ – Michael Rozenberg Jan 15 '18 at 3:52

The task is homogenius and symmetric by the $b,c.$

Let WLOG $$a=1,\quad s = \dfrac{b+c}2,\quad t=\left(\dfrac{b-c}{b+c}\right)^2,\tag1$$ then \begin{align} &b+c = 2s,\quad bc = s^2(1-t),\quad b^2+c^2=2s^2(1+t),\\[4pt] &b^4+c^4 = (b^2+c^2)^2 - 2(bc)^2 = 2s^4(t^2+6t+1),\\[4pt] &\dfrac{b}{1+c}+\dfrac{c}{1+b} = \dfrac{b+c+b^2+c^2}{1+b+c+bc} = \dfrac{2s+2s^2(1+t)}{1+2s+s^2(1-t)}, \end{align} and the issue task transforms to $$\dfrac1{s\sqrt[4]{t^2+6t+1}} + \dfrac{4s+4s^2(1+t)}{1+s+s^2(1-t)} \ge 3, \quad s > 0,\quad 1\ge t\ge0,$$ or $$\dfrac1{s\sqrt[4]{t^2+6t+1}} + 4\dfrac{(1+s)(1+2s)}{(1+s)^2-s^2t} \ge 7, \quad s > 0,\quad 1\ge t\ge0,$$ or $$\dfrac{v-1}{\sqrt[4]{t^2+6t+1}} + 4\dfrac{v(v+1)}{v^2-t} \ge 7, \quad v > 1,\quad 1\ge t\ge0, \tag2,$$ where $$v=\dfrac1s+1.\tag3$$ Then, using the evident inequality $$\sqrt{1\pm x}\le 1\pm \dfrac x2,$$ easy to get $$\sqrt{1+6t^2+t^4} = (1+3t)\sqrt{1-\dfrac {8t^2}{(1+3t)^2}}\le(1+3t)\left(1-\dfrac{4t^2}{(1+3t)^2}\right) = \dfrac{1+6t+5t^2}{1+3t},$$ $$\sqrt{1+6t+5t^2} = (1+3t)\sqrt{1-\dfrac {4t^2}{(1+3t)^2}}\le\dfrac{1+3t}{1+\dfrac{2t^2}{(1+3t)^2}} = \dfrac{(1+3t)^3}{1+6t+11t^2}.$$

This allows to prove the stronger inequality $${1+6t+11t^2\over (\sqrt{1+3t\,})^5}(v-1) + 4\dfrac{v(v+1)}{v^2-t} \ge 7, \quad v > 1,\quad 1\ge t\ge0.$$ Substitution $$z=\sqrt{1+3t}\tag4$$ leads to the equivalent task $${2-4z^2+11z^4\over 9z^5}(v-1)+12{v(v+1)\over 3v^2 + 1-z^2} \ge 7,\quad z\ge 1,\quad 2 \ge z\ge 1,$$ or $$f(z,v) \ge 0,\quad v>1,\quad 2 \ge z\ge1,\tag5$$ wherein \begin{align} f(z,v) = (3v^2+1-z^2)((v-1)(11z^4-4z^2+2)-63z^5)+108z^5(v^2+v).\tag6 \end{align}

The highest value of $f(x)$ can be achieved on its stationary points or the area bounds.

The stationary points

The stationary points of $f(z,v)$ can be obtained from the conditions $f'_v=\dfrac1{3z}f'_v=0,$ and that leads to the system

\begin{cases} (99z^4-36z^2+18)v^2+(-162z^5-66z^4+24z^2-12)v-11z^6+108z^5+15z^4-6z^2+2 = 0\\[4pt] (44z^2-8)v^3+(-135z^3-44z^2+8)v^2+(-22z^4+180z^3+20z^2-4)v+147z^5+22z^4-105z^3-20z^2+4=0\\[4pt] v>1,\quad 2 \ge z\ge1,\tag7 \end{cases} (see also Wolfram Alpha), with the solutions

$$\begin{pmatrix}v\\z\\f(v,z)\end{pmatrix} =\left\{ \begin{pmatrix}2\\1\\0\end{pmatrix}, \begin{pmatrix}\approx 1.713140\\\approx 0.883476\\ \approx 0.0969191\end{pmatrix} \right\}\tag8$$ which can be obtained using the polynomial reduction way.

Taking in account the calculations accuracy, this means that $f(v,z)\ge 0$ in the stationary points.

The bounds

If $\mathbf{v=1,}$ then $f(1,z) = 9z^5(7z^2-4)\ge 0\text{ if } z\in[1,2].$

If $\mathbf{z=1},$ then $f(v, 1) = 27v(v-2)^2 \ge 0 \text{ if } v\in(1,\infty).$

If $\mathbf{z=2},$ then $f(v,2) = 54 (11 - 3 v)^2 (v + 1) \ge 0,\text{ if } v\in(1,\infty).$

These mean that $f(v,z)\ge 0$ on the area $v\in(1, \infty), \quad z\in [1,2].$


Update 08.05.2018

The polynomial reduction way

Taking in account the elder coefficients of the derivatives, let us consider the system $$\dfrac{df}{dv} = 4v(11z^2-2)\dfrac{df}{dv} - \dfrac{-33z^4+12z^2-6}{z}\dfrac{df}{dz} = 0,$$ or $$-6(27z^5+11z^4-4z^2+2)v+9(11z^4-4z^2+2)v^2 = 11z^6-108z^5-15z^4+6z^2-2\\ 2(847z^8-6534z^7-1012z^6+2808z^5+564z^4-1620z^3-184z^2+28)v+ 3(2079z^7+484z^6-1188z^5-264z^4+810z^3+120z^2-16)v^2 = 9(1617z^9+242z^8-1743z^7-308z^6+714z^5+168z^4-210z^3-56z^2+8) $$ $$\begin{cases} -6(27z^5+11z^4-4z^2+2)v + 9(11z^4-4z^2+2)v^2=11z^6-108z^5-15z^4+6z^2-2\\[4pt] 2(847z^8-6534z^7-1012z^6+2808z^5+564z^4-1620z^3-184z^2+28)v\\ +3(2079z^7+484z^6-1188z^5-264z^4+810z^3+120z^2-16)v^2\\ =9(1617z^9+242z^8-1743z^7-308z^6+714z^5+168z^4-210z^3-56z^2+8)\\[4pt] v>1,\quad 2 \ge z\ge1, \end{cases}\tag9$$ (see also Wolfram Alfa) which can consist the additional roots besides the roots of $(7).$ Consideraton of $(9)$ as the linear by $\{s, s^2\}$ leads to the solution $$\begin{pmatrix} v\\ v^2\end{pmatrix} = \dfrac1{\Delta}\begin{pmatrix} \Delta_1\\ \Delta_2\end{pmatrix},\tag{10}$$ where \begin{align} \Delta&= \begin{vmatrix} -6(27z^5+11z^4-4z^2+2)&9(11z^4-4z^2+2)\\[4pt] \genfrac{.}{.}{0}{0}{2(847z^8-6534z^7-1012z^6+2808z^5}{+564z^4-1620z^3-184z^2+28)} &\genfrac{.}{.}{0}{0}{3(2079z^7+484z^6}{-1188z^5-264z^4+810z^3+120z^2-16)} \end{vmatrix}\\[4pt] &=-18(65450z^{12}-35937z^{11}-41272z^{10}+28512z^9+28976z^8-21060z^7\\ &-2960z^6+6048z^5+988z^4-1620z^3-176z^2+24),\\ \end{align} $$\Delta = -18d(z),\tag{11}$$ $$\begin{align} d&=65450z^{12}-35937z^{11}-41272z^{10}+28512z^9+28976z^8-21060z^7\\ &-2960z^6+6048z^5+988z^4-1620z^3-176z^2+24, \end{align}\tag{12}$$ \begin{align} \Delta_1&= \begin{vmatrix} 11z^6-108z^5-15z^4+6z^2+2&9(11z^4-4z^2+2)\\[4pt] \genfrac{.}{.}{0}{0}{9(1617z^9+242z^8-1743z^7-308z^6}{+714z^5+168z^4-210z^3-56z^2+8)} &\genfrac{.}{.}{0}{0}{3(2079z^7+484z^6-1188z^5}{-264z^4+810z^3+120z^2-16)} \end{vmatrix}\\[4pt] &=-6(228690z^{13}+145541z^{12}-297891z^{11}-117876z^{10}+209952z^9+87762z^8\\ &-102762z^7-24408z^6+28512z^5+8532z^4-6480z^3-2016z^2+232),\\ \end{align} $$\Delta_1=-6d_1(z),\tag{13}$$ $$\begin{align} d_1(z)&=228690z^{13}+145541z^{12}-297891z^{11}-117876z^{10}+209952z^9+87762z^8\\ &-102762z^7-24408z^6+28512z^5+8532z^4-6480z^3-2016z^2+232, \end{align}\tag{14}$$ \begin{align} \Delta_2&= \begin{vmatrix} -6(27z^5+11z^4-4z^2+2)&11z^6-108z^5-15z^4+6z^2+2\\[4pt] \genfrac{.}{.}{0}{0}{2(847z^8-6534z^7-1012z^6+2808z^5}{+564z^4-1620z^3-184z^2+28)} &\genfrac{.}{.}{0}{0}{9(1617z^9+242z^8-1743z^7-308z^6}{+714z^5+168z^4-210z^3-56z^2+8)} \end{vmatrix},\\[4pt] &=-2(1188110z^{14}+493317z^{13}-516938z^{12}-678645z^{11}+126096z^{10}+450036z^9\\ &+103236z^8-226476z^7-46980z^6+59940z^5+17100z^4-14580z^3-4088z^2+488).\\ \end{align} $$\Delta_2 = -2d_2(z),\tag{15}$$ $$\begin{align} d_2(z)&=1188110z^{14}+493317z^{13}-516938z^{12}-678645z^{11}+126096z^{10}+450036z^9\\ &+103236z^8-226476z^7-46980z^6+59940z^5+17100z^4-14580z^3-4088z^2+488,\\ \end{align}\tag{16}$$ $(10),(12),(14),(16)$ allow to write $$v=\dfrac{d_1(z)}{3d(z)},\quad d_1(z)^2 = d_2(z)d(z),$$ or $$ \begin{cases} 76977054000z^{25}-25462683400z^{26}-14469601850z^{24}-128299931694z^{23}+\\ 47986668444z^{22}+144016029432z^{21}-67667920532z^{20}-106475101020z^{19}+\\ 50533574480z^{18}+61655670720z^{17}-26340071868z^{16}-27611975304z^{15}+\\ 9481920960z^{14}+10104266592z^{13}-2350471584z^{12}-3006244368z^{11}+331237872z^{10}+\\721643904z^9+10846384z^8-136453248z^7-19240960z^6+\\ 17418240z^5+5038080z^4-1140480z^3-563776z^2+30976=0\\ v=\dfrac{d_1(z)}{3d(z)}, v>1,\quad 2 \ge z\ge1. \end{cases}\tag{17}$$ The first equation allows partial decomposition to the form of $$\begin{align} &(z-1)(11z^4-4z^2+2)\times\\ &(1157394700z^{21}-2341562300z^{20}-1262982325z^{19}+3296484752z^{18} +1297048050z^{17}\\ &-2954973078z^{16}-802331850z^{15}+1662261954z^{14}+511593870z^{13}-621625938z^{12}\\ &-234164016z^{11}+160737876z^{10}+79849092z^9-29795100z^8-22817208z^7+2159304z^6\\ &+4347504z^5+563184z^4-402832z^3-117712z^2+7744z+7744), \end{align}\tag{18}$$ where the last polynomial factor has the roots $$\begin{align} &z\in\{-0.65254852+0.411596443i,\quad -0.65254852-0.411596443i,\quad -0.533266085,\\ &-0.463163014+0.296052158i,\quad -0.463163014-0.296052158i,\\ &-0.423458642-0.209546484i,\quad -0.423458642+0.209546484i,\\ &-0.242560904-0.338675996i,\quad -0.242560904+0.338675996i,\\ &-0.220398679-0.318470135i,\quad -0.220398679+0.318470135i,\\ &0.334079252+0.005204691i,\quad 0.334079252-0.005204691i,\\ &0.397903595+0.409264586i,\quad 0.397903595-0.409264586i,\\ &0.553362795-0.024077867i,\quad 0.553362795+0.024077867i,\\ &0.658848407+0.496670038i,\quad 0.658848407-0.496670038i,\\ &0.883475687,\quad 1.788793855\}.\\ \end{align}$$ Easy to see that this leads to the result $(8).$

  • $\begingroup$ If you made the slightest mistake, everything would be wrong. $(+1)$ $\endgroup$ – Feeds May 8 '18 at 0:13
  • $\begingroup$ @user477343 I've tried to simplify the equation further, but the coefficients only increased. $\endgroup$ – Yuri Negometyanov May 8 '18 at 0:23
  • $\begingroup$ I'm impressed by the amount of work, whether or not there is an error. $\endgroup$ – marty cohen Jun 10 '18 at 19:08

Your inequality is equivalent to : $$f(x)=\frac{1}{(8(x^4+h^4))^{0.25}}+\frac{x}{h+1}+\frac{h}{x+1}$$ Here we assume that $x\geq h \geq 1$

It's easy to see that the function $f(x)$ is convex with $h$ fixed (as the sum of convexs functions) and increasing for $x\geq h$ Furthermore for $h$ fixed we have that the minimum is inferior to $h$ (in abscissa)

So we have :

$$f(x)\geq f(h)$$

We get : $$f(x)=\frac{1}{(8(x^4+h^4))^{0.25}}+\frac{x}{h+1}+\frac{h}{x+1}\geq\frac{1}{(8(h^4+h^4))^{0.25}}+\frac{h}{h+1}+\frac{h}{h+1} $$

So the inequality is verified in this case because it's easy to threat this one variable inequality.

The case where $h\geq x \geq 1$ is the same (because of the symetry) and the cases $x\geq 1 \geq h$ and

$h\geq 1 \geq x$ are similar for the same reasons

Now we continue with the case $x\leq h \leq 1$

Remark that if we inverse each variable it becomes :


With $h\geq 1$ and $x\geq 1$ but we have :

$$\frac{xh}{(8(x^4+h^4))^{0.25}}+\frac{x}{h+1}+\frac{h}{x+1}\geq \frac{1}{(8(x^4+h^4))^{0.25}}+\frac{x}{h+1}+\frac{h}{x+1}\geq 1.5$$

So this case is verified .

Done !

Edit :My proof is incomplete like this because the case where $x\geq 1 \geq h$ is not solved . So I purpose to you a solution :

The main idea is to use the three chord lemma related to the convexity of the function $f(x)$ so we have for $x\geq \sqrt{x}\geq x^{0.25}$ :

$$\frac{f(x)-f(\sqrt{x})}{x-\sqrt{x}}\geq \frac{f(x)-f(x^{0.25})}{x-x^{0.25}} \geq \frac{f(\sqrt{x})-f(x^{0.25})}{\sqrt{x}-x^{0.25}}$$

We can apply this inequality $n$ times to get :

$$\frac{f(x)-f(\sqrt{x})}{x-\sqrt{x}}\geq \frac{f(x^{1/2^n})-f(x^{1/2^{n+1}})}{x^{1/2^n}-x^{1/2^{n+1}}}$$ But we have to prove that the RHS is positive for a variable wich tends to $1$ so we get the following result to evaluate :

$$\lim_{x\to 1} \frac{\dfrac{1}{(8(x^{1/2^n}+h^4))^{0.25}}+\dfrac{x^{1/2^n}}{h+1} + \dfrac{h}{x^{1/2^n}+1}- \left(\dfrac{1}{(8(x^{1/2^{n+1}}+h^4))^{0.25}} + \dfrac{x^{1/2^{n+1}}}{h+1}+\dfrac{h}{x^{1/2^{n+1}}+1} \right)}{x^{1/2^n}-x^{1/2^{n+1}}}$$

For that see the the interesting answer of G.Cab here

With this answer we know that the result does not depends on $n$ and is equal to :

$$ \eqalign{ & \mathop { } & {1 \over {h + 1}} - {1 \over {2^{\,11/4} \left( {h^{\,4} + 1} \right)^{\,5/4} }} - {h \over 4} \cr} $$

Quantity wich is positive for $h\leq 1$

So we get :

$$\frac{f(x)-f(\sqrt{x})}{x-\sqrt{x}}\geq 0$$

Or :

$$f(x)\geq f(\sqrt{x})$$

If we apply this inequality $n$ times we get :

$$f(x)\geq f(x^{\frac{1}{2^n}})$$

But the variable tends $1$ and the RHS to $1.5$ so we get the desired result !

  • $\begingroup$ @max1828 Does it work for n=5? =) $\endgroup$ – Yuri Negometyanov May 1 '18 at 15:37
  • $\begingroup$ Yes it works with the coefiicient $8$ why ? There is a mistake in my reasoning ? $\endgroup$ – max8128 May 1 '18 at 17:07
  • $\begingroup$ @max1828 Inequality with $n=5$ and coefficient $16$ should be wrong $\endgroup$ – Yuri Negometyanov May 1 '18 at 17:51
  • $\begingroup$ See attentively my answer . With the inequality withe one variable we can see that it works for large value . I will post a question about this .Thanks for your interest ! $\endgroup$ – max8128 May 2 '18 at 8:38
  • $\begingroup$ The proposed "proof" is erroneous. Try to find the error yourself. $\endgroup$ – Yuri Negometyanov May 3 '18 at 8:48

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