Stronger than Nesbitt's inequality $\frac{a}{\sqrt[4]{8(b^4+c^4)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}$

Solution 1:

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 $v\in(1, \infty), \quad z\in [1,2].$

Proved.

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 $(8).$

Solution 2:

With computer, here is a solution.

WLOG, assume that $b \ge c$.

(1) If $c = 0$, we need to prove that $$\frac{a}{2^{3/4}b} + \frac{b}{a} \ge \frac{3}{2}.$$ Using AM-GM, we have $\frac{a}{2^{3/4}b} + \frac{b}{a} \ge 2\sqrt{\frac{a}{2^{3/4}b} \frac{b}{a}} = 2^{5/8} > \frac{3}{2}$. The inequality is true.

(2) If $c > 0$, WLOG, assume that $c=1$. Let $b = 1 + x, \ x \ge 0$. We need to prove that $$\frac{a}{\sqrt[4]{8((1+x)^4+1)}} + \frac{1+x}{a+1} + \frac{1}{a+1+x} \ge \frac{3}{2}.$$

We will use the following bound: $$\frac{4(1+x)^2(165x^2+144x+112)}{389x^3+734x^2+624x+224} \ge \sqrt[4]{8((1+x)^4+1)}, \quad \forall x\ge 0$$ which follows from \begin{align} &\Big(\frac{4(1+x)^2(165x^2+144x+112)}{389x^3+734x^2+624x+224}\Big)^4 - 8((1+x)^4+1)\\ =\ & \frac{8x^6}{(389x^3+734x^2+624x+224)^4}\\ &\times \left(820374959x^{10}+8129708452x^9+34536522338x^8+88010431212x^7\right.\\ &\quad +150840675678x^6+183012461808x^5 + 160216957296x^4+100447002624x^3\\ &\quad \left. +43418651136x^2+11761455104x+1518727168\right). \end{align} Thus, it suffices to prove that $$\frac{a(389x^3+734x^2+624x+224)}{4(1+x)^2(165x^2+144x+112)} + \frac{1+x}{a+1} + \frac{1}{a+1+x} \ge \frac{3}{2}.$$ After clearing the denominators, it suffices to prove that $$q_3a^3 + q_2a^2 + q_1 a + q_0 \ge 0$$ where \begin{align} q_3 &= 389 x^3+734 x^2+624 x+224\\ q_2 &= -601 x^4-1332 x^3-1298 x^2-736 x-224\\ q_1 &= -330 x^5-1219 x^4-1903 x^3-1638 x^2-848 x-224, \\ q_0 &= 660 x^6+2226 x^5+3538 x^4+3550 x^3+2314 x^2+960 x+224. \end{align}

If $x=0$, we have $q_3a^3 + q_2a^2 + q_1 a + q_0 = 224(a+1)(a-1)^2 \ge 0$. True.

If $x > 0$, let $f(u) = q_3u^3 + q_2u^2 + q_1 u + q_0$. The discriminant of $f(u)$ is \begin{align} &\mathrm{discr}(f) = -(1+x)^2x^4 \\ &\times \left(194821434060 x^{12}+1160408972196 x^{11}+3444683967175 x^{10}+6771294350288 x^9 \right.\\ &\quad +9755714292752 x^8+10819604233984 x^7+9460649716736 x^6+6565740839936 x^5\\ &\quad +3594172926976 x^4+1515569807360 x^3+468531445760 x^2\\ &\quad \left. +95759892480 x+9441116160\right). \end{align} Since $\mathrm{discr}(f) < 0$, the equation $f(u)=0$ has only one real root. Note also that $f(0) = q_0 > 0$ and $f(-\infty) = -\infty$ (since $q_3>0$). Thus, the only one real root of $f(u)=0$ is negative. Thus, $f(a) > 0$ for $a \ge 0$. We are done.