Prove : $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}}$

Question

It's an inequality based on two found on the website MSE (see the reference):

Let $a,b,c>0$ then we have:

$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}}$$

Lemma 1 :

$a,b,c>0$ then we have :

$$\sum_{cyc}\frac{a}{b+c}\geq P(a,b,c)=\sqrt{\frac{9}{4}+\frac{9}{4}\frac{(c^4+a^4+b^4-c^2a^2-b^2a^2-c^2b^2)}{((a+b+c)\frac{3}{4}+\frac{3}{4}(abc)^{\frac{1}{3}})(a+b)(b+c)(c+a)}+\frac{(c^2+a^2+b^2-ca-ba-cb)(a+b+c)}{(a+b)(b+c)(c+a)}}$$

Proof of lemma 1 :

First we remark that the inequality is homogenous and we can try the substitution $3u=a+b+c$, $3v^2=ab+bc+ca$ and $w^3=abc$ and apply the uvw's method .

We have : $$a^4+b^4+c^4=(9u^2-6v^2)^2-2(9v^4-6uw^3)$$ $$a^2b^2+b^2c^2+c^2a^2=9v^4-6uw^3$$ $$a^2+b^2+c^2=9u^2-6v^2$$

And : $\left(\frac{((3u)((3u)^2-4(3v^2))+5w^3)}{3u3v^2-w^3}+2\right)^2\geq \frac{9}{4}+\frac{(9/4)((9u^2-6v^2)^2-2(9v^4-6uw^3)-(9v^4-6uw^3))+(2.25u+0.75w)(9u^2-9v^2)(3u)}{(2.25u+0.75w)(3u3v^2-w^3)}$

it's enough to find an extreme value of our expression for the extreme value of $w^3$ wich happens for an equality case of two variables .

Since the last inequality is homogeneous, we can assume that $b=c=1$.

$$\frac{2}{a+1}+\frac{a}{2}\geq\sqrt{\frac{9}{4}+\frac{9}{4}\frac{(a^4+1-2a^2)}{((a+2)\frac{3}{4}+\frac{3}{4}(a)^{\frac{1}{3}})(a+1)^2(2)}+\frac{(a^2+1-2a)(a+2)}{(a+1)^2(2)}}$$

Now it seems to be clear : we get a polynomial with a root equal to one . See the factorization by Wolfram alpha .

End of the proof of the lemma 1

Remains to show that $ a\geq b \geq c>0$:

$$P(a,b,c)\geq\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}}$$

Wich is not hard I think .

Question :

How to show it ?

Reference :

M. A. Rozenberg, “uvw–Method in Proving Inequalities”, Math. Ed., 2011, no. 3-4(59-60), 6–14

If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$

Stronger than Nesbitt inequality

Answer 1

Let $a,b,c>0$ then we have:

$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq \sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}}$$ say that $x \in \{ a,b,c \}$ then $x^3-px^2+qx-r = 0$ $$p = a+b+c$$ $$q = ab+ac+bc$$ $$r= abc$$ $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} = \frac{ p^3-2pq+3r}{pq-r} $$ $$\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}} = \sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2p}{pq-r}}$$ $$\frac{ p^3-2pq+3r}{pq-r} \geq \sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2p}{pq-r}}$$ $(a-b)^2$ is not symmetric, so it is the root of some cubic in $y$, with its conjugate being $(a-c)^2$ and $(b-c)^2$ $$y^3+(6q-2p^2)y^2+(9q^2-6p^2q+p^4)y+27r^2-18pqr+4p^3r+4q^3-p^2q^2 = 0$$ $$\frac{ p^3-2pq+3r}{pq-r} \geq \sqrt{\frac{9}{4}+\frac{3}{2}\frac{py}{pq-r}}$$ our trick here is to make the RHS a root of another polynomial $z$ $$\frac{ p^3-2pq+3r}{pq-r} \geq \sqrt{z}$$ $$( \frac{ p^3-2pq+3r}{pq-r} )^2 \geq z$$ $$ z = \frac{9}{4}+\frac{3}{2}\frac{py}{pq-r}$$ solving for $y$ so that we can substitute it's value to get the expression for polynomial $z$ $$ y = \frac{ (4z-9)(pq-r) }{6p}$$ $$y^3+(6q-2p^2)y^2+(9q^2-6p^2q+p^4)y+27r^2-18pqr+4p^3r+4q^3-p^2q^2 = 0$$ $$ (64r^3-192pqr^2+192p^2q^2r-64p^3q^3)z^3+(-432r^3+720pqr^2+192p^3r^2-144p^2q^2r-384p^4qr-144p^3q^3+192p^5q^2)z^2+(972r^3-324pqr^2-864p^3r^2-972p^2q^2r+864p^4qr+144p^6r+324p^3q^3-144p^7q)z-729r^3-729pqr^2-4860p^3r^2+729p^2q^2r+3888p^4qr-1188p^6r-135p^3q^3-756p^5q^2+324p^7q = 0$$ from $( \frac{ p^3-2pq+3r}{pq-r} )^2 \geq z$ since the inequality here is greater than our root $z$, if I put this for $z$ the value is expected to be also greater than or equal to $0$ $$ (64r^3-192pqr^2+192p^2q^2r-64p^3q^3)\cdot (\frac{ p^3-2pq+3r}{pq-r} )^6+(-432r^3+720pqr^2+192p^3r^2-144p^2q^2r-384p^4qr-144p^3q^3+192p^5q^2)\cdot ( \frac{ p^3-2pq+3r}{pq-r} )^4+(972r^3-324pqr^2-864p^3r^2-972p^2q^2r+864p^4qr+144p^6r+324p^3q^3-144p^7q)\cdot ( \frac{ p^3-2pq+3r}{pq-r} )^2-729r^3-729pqr^2-4860p^3r^2+729p^2q^2r+3888p^4qr-1188p^6r-135p^3q^3-756p^5q^2+324p^7q \geq 0$$ $$\frac{ 19683r^6-91854pqr^5+55404p^3r^5+181521p^2q^2r^4-210924p^4qr^4+71064p^6r^4-193428p^3q^3r^3+331128p^5q^2r^3-210816p^7qr^3+39744p^9r^3+117045p^4q^4r^2-265752p^6q^3r^2+241200p^8q^2r^2-86976p^{10}qr^2+10656p^{12}r^2-38142p^5q^5r+108444p^7q^4r-125376p^9q^3r+65088p^{11}q^2r-15360p^{13}qr+1344p^{15}r+5239p^6q^6-17916p^8q^5+24888p^{10}q^4-16576p^{12}q^3+5664p^{14}q^2-960p^{16}q+64p^{18}}{(r-pq)^3} \geq 0$$ $$19683r^6-91854pqr^5+55404p^3r^5+181521p^2q^2r^4-210924p^4qr^4+71064p^6r^4-193428p^3q^3r^3+331128p^5q^2r^3-210816p^7qr^3+39744p^9r^3+117045p^4q^4r^2-265752p^6q^3r^2+241200p^8q^2r^2-86976p^{10}qr^2+10656p^{12}r^2-38142p^5q^5r+108444p^7q^4r-125376p^9q^3r+65088p^{11}q^2r-15360p^{13}qr+1344p^{15}r+5239p^6q^6-17916p^8q^5+24888p^{10}q^4-16576p^12q^3+5664p^{14}q^2-960p^{16}q+64p^{18} \geq (r-pq)^3$$ now for $a,b,c \gt 0$, we are left to prove the above inequality, obviously without calculating I can see how great the LHS is, compared to the RHS..... also since $a,b,c \gt 0$, it is easy to know that $q \gt ( p, r)$ and sometimes $q \gt r \gt p $, this makes me suspect that the expression $r-pq$ is negative

If I write in terms of the original elements $a,b,c$ then $$64c^{18}+192bc^{17}+192ac^{17}+96b^2c^{16}+576abc^{16}+96a^2c^{16}-256b^3c^{15}-256a^3c^{15}-360b^4c^{14}-1248ab^3c^{14}-816a^2b^2c^{14}-1248a^3bc^{14}-360a^4c^{14}-204b^5c^{13}-924ab^4c^{13}-792a^2b^3c^{13}-792a^3b^2c^{13}-924a^4bc^{13}-204a^5c^{13}-369b^6c^{12}-630ab^5c^{12}+1281a^2b^4c^{12}+4364a^3b^3c^{12}+1281a^4b^2c^{12}-630a^5bc^{12}-369a^6c^{12}-390b^7c^{11}-1626ab^6c^{11}+690a^2b^5c^{11}+7470a^3b^4c^{11}+7470a^4b^3c^{11}+690a^5b^2c^{11}-1626a^6bc^{11}-390a^7c^{11}+681b^8c^{10}+138ab^7c^{10}-2736a^2b^6c^{10}+630a^3b^5c^{10}+6606a^4b^4c^{10}+630a^5b^3c^{10}-2736a^6b^2c^{10}+138a^7bc^{10}+681a^8c^{10}+1540b^9c^9+4098ab^8c^9-90a^2b^7c^9-6566a^3b^6c^9-5382a^4b^5c^9-5382a^5b^4c^9-6566a^6b^3c^9-90a^7b^2c^9+4098a^8bc^9+1540a^9c^9+681b^{10}c^8+4098ab^9c^8+3966a^2b^8c^8-6162a^3b^7c^8-8439a^4b^6c^8-3072a^5b^5c^8-8439a^6b^4c^8-6162a^7b^3c^8+3966a^8b^2c^8+4098a^9bc^8+681a^{10}c^8-390b^{11}c^7+138ab^{10}c^7-90a^2b^9c^7-6162a^3b^8c^7-4152a^4b^7c^7+11424a^5b^6c^7+11424a^6b^5c^7-4152a^7b^4c^7-6162a^8b^3c^7-90a^9b^2c^7+138a^{10}bc^7-390a^{11}c^7-369b^{12}c^6-1626ab^{11}c^6-2736a^2b^{10}c^6-6566a^3b^9c^6-8439a^4b^8c^6+11424a^5b^7c^6+29616a^6b^6c^6+11424a^7b^5c^6-8439a^8b^4c^6-6566a^9b^3c^6-2736a^{10}b^2c^6-1626a^{11}bc^6-369a^{12}c^6-204b^{13}c^5-630ab^{12}c^5+690a^2b^{11}c^5+630a^3b^{10}c^5-5382a^4b^9c^5-3072a^5b^8c^5+11424a^6b^7c^5+11424a^7b^6c^5-3072a^8b^5c^5-5382a^9b^4c^5+630a^{10}b^3c^5+690a^{11}b^2c^5-630a^{12}bc^5-204a^{13}c^5-360b^{14}c^4-924ab^{13}c^4+1281a^2b^{12}c^4+7470a^3b^{11}c^4+6606a^4b^{10}c^4-5382a^5b^9c^4-8439a^6b^8c^4-4152a^7b^7c^4-8439a^8b^6c^4-5382a^9b^5c^4+6606a^{10}b^4c^4+7470a^{11}b^3c^4+1281a^{12}b^2c^4-924a^{13}bc^4-360a^{14}c^4-256b^{15}c^3-1248ab^{14}c^3-792a^2b^{13}c^3+4364a^3b^{12}c^3+7470a^4b^{11}c^3+630a^5b^{10}c^3-6566a^6b^9c^3-6162a^7b^8c^3-6162a^8b^7c^3-6566a^9b^6c^3+630a^{10}b^5c^3+7470a^{11}b^4c^3+4364a^{12}b^3c^3-792a^{13}b^2c^3-1248a^{14}bc^3-256a^{15}c^3+96b^{16}c^2-816a^2b^{14}c^2-792a^3b^{13}c^2+1281a^4b^{12}c^2+690a^5b^{11}c^2-2736a^6b^{10}c^2-90a^7b^9c^2+3966a^8b^8c^2-90a^9b^7c^2-2736a^{10}b^6c^2+690a^{11}b^5c^2+1281a^{12}b^4c^2-792a^{13}b^3c^2-816a^{14}b^2c^2+96a^{16}c^2+192b^{17}c+576ab^{16}c-1248a^3b^{14}c-924a^4b^{13}c-630*a^5b^{12}c-1626a^6b^{11}c+138a^7b^{10}c+4098a^8b^9c+4098a^9b^8c+138a^{10}b^7c-1626a^{11}b^6c-630a^{12}b^5c-924a^{13}b^4c-1248a^{14}b^3c+576a^{16}bc+192a^{17}c+64b^{18}+192ab^{17}+96a^2b^{16}-256a^3b^{15}-360a^4b^{14}-204a^5b^{13}-369a^6b^{12}-390a^7b^{11}+681a^8b^{10}+1540a^9b^9+681a^{10}b^8-390a^{11}b^7-369a^{12}b^6-204a^{13}b^5-360a^{14}b^4-256a^{15}b^3+96a^{16}b^2+192a^{17}b+64a^{18} \geq -bc^2-ac^2-b^2c-2abc-a^2c-ab^2-a^2b$$

Just as I said the RHS is negative, we are going to prove that the LHS is positive, though it is symmetric and it factors into $3$ products $$(4c^6+2bc^5+2ac^5-9b^2c^4+2abc^4+3a^2c^4-8b^3c^3-14ab^2c^3+4a^2bc^3+10a^3c^3+3b^4c^2-8ab^3c^2-6a^2b^2c^2+4a^3bc^2+3a^4c^2+8b^5c+14ab^4c-8a^2b^3c-14a^3b^2c+2a^4bc+2a^5c+4b^6+8ab^5+3a^2b^4-8a^3b^3-9a^4b^2+2a^5b+4a^6)\cdot (4c^6+2bc^5+2ac^5+3b^2c^4+2abc^4-9a^2c^4+10b^3c^3+4ab^2c^3-14a^2bc^3-8a^3c^3+3b^4c^2+4ab^3c^2-6a^2b^2c^2-8a^3bc^2+3a^4c^2+2b^5c+2ab^4c-14a^2b^3c-8a^3b^2c+14a^4bc+8a^5c+4b^6+2ab^5-9a^2b^4-8a^3b^3+3a^4b^2+8a^5b+4a^6)\cdot (4c^6+8bc^5+8ac^5+3b^2c^4+14abc^4+3a^2c^4-8b^3c^3-8ab^2c^3-8a^2bc^3-8a^3c^3-9b^4c^2-14ab^3c^2-6a^2b^2c^2-14a^3bc^2-9a^4c^2+2b^5c+2ab^4c+4a^2b^3c+4a^3b^2c+2a^4bc+2a^5c+4b^6+2ab^5+3a^2b^4+10a^3b^3+3a^4b^2+2a^5b+4a^6) \geq -bc^2-ac^2-b^2c-2abc-a^2c-ab^2-a^2b$$ mathematical speaking, by looking at each product there it's easy to know the the inequality has been solved

but for op sake, since the LHS is positive, therefore it's three products are all positive too and it turns out these products are almost symmetric ( symmetric in $2$ elements, $x_1 \to x_2$), now randomly selecting an order $$4c^6+2bc^5+2ac^5-9b^2c^4+2abc^4+3a^2c^4-8b^3c^3-14ab^2c^3+4a^2bc^3+10a^3c^3+3b^4c^2-8ab^3c^2-6a^2b^2c^2+4a^3bc^2+3a^4c^2+8b^5c+14ab^4c-8a^2b^3c-14a^3b^2c+2a^4bc+2a^5c+4b^6+8ab^5+3a^2b^4-8a^3b^3-9a^4b^2+2a^5b+4a^6 \geq 0 $$ I selected one because the three are related and are all positive, If $a,b,c$ are equal this expression is $0$, if $a,b,c \geq 0$ it is true $$ 4c^6+2bc^5+2ac^5+2abc^4+3a^2c^4+4a^2bc^3+10a^3c^3+3b^4c^2+4a^3bc^2+3a^4c^2+8b^5c+14ab^4c+2a^4bc+2a^5c+4b^6+8ab^5+3a^2b^4+2a^5b+4a^6 \geq 9b^2c^4+8b^3c^3+14ab^2c^3+8ab^3c^2+6a^2b^2c^2+8a^2b^3c+14a^3b^2c+8a^3b^3+9a^4b^2 $$ $$ 4c^6+2bc^5+2ac^5+2abc^4+3a^2c^4+4a^2bc^3+10a^3c^3+3b^4c^2+4a^3bc^2+3a^4c^2+8b^5c+14ab^4c+2a^4bc+2a^5c+4b^6+8ab^5+3a^2b^4+2a^5b+4a^6 \geq b^2(9c^4+8bc^3+14ac^3+8abc^2+6a^2c^2+8a^2bc+14a^3c+8a^3b+9a^4) $$ by comparing the weights of the RHS and LHS, the inequality is true

Answer 2

The given inequality is homogenious by $\;(a,b,c)\;$ and symmetric by $\;(a,b).\;$

Let WLOG $$a+b+c = 6,\quad a+b=s,\quad ab=p,\tag1$$ then it suffices to prove the inequality in the form of $$\dfrac a{6-a}+\dfrac b{6-b}+\dfrac{6-a-b}{a+b} \ge \sqrt{\dfrac94 +9\,\dfrac{(a-b)^2}{(a+b)(6-a)(6-b)}},\tag2$$ with the both positive sides, wherein the square root function monotonically increases in $\;[0,\infty).\;$

Therefore, one may to square both of the inequality sides and eliminate the positive denominators, with the represntations \begin{align} &\bigg(\big(a(6-b)+b(6-a)\big)(a+b)+(6-a-b)(6-a)(6-b)\bigg)^2\\[4pt] &\ge \dfrac94\bigg(\big((6-a)(6-b)(a+b)\big)^2-4(a-b)^2(6-a)(6-b)(a+b)\bigg), \tag3 \end{align} or $$\dfrac49\big(6s^2-2ps + 6(6-s)^2+p(6-s)\big)^2$$ $$\ge (36s-6s^2+ps)^2-4(s^2-4p)(36s-6s^2+ps),$$ or $$f(s,p)\ge 0,\tag4$$ where $$s^2\ge4p,\quad 6\ge s,\tag5$$ $$f(s,p)=\big(144-48s+8s^2+4p-2ps\big)^2 - (36s-8s^2+8p+ps)^2+4(s^2-4p)^2$$ $$=\big(144-84s+16s^2-4p-3ps\big)\big(144-12s+12p-ps\big)+4(s^2-4p)^2$$ $$\ge\dfrac34\big(192-112s+20s^2-s^3\big)(12-s)(12+p)$$ $$=\dfrac34(s-4)^2(12-s)^2(12+p)\ge0.$$ Proved!

Answer 3

As username says it's not symmetric so here there is a refinement and a symmetric formula :

$a,b,c\in[0.95,1]$:

$$\sum_{cyc}\frac{a}{b+c}\geq\sqrt{\frac{9}{4}+\frac{9}{4}\frac{(c^4+a^4+b^4-c^2a^2-b^2a^2-c^2b^2)}{(0.75(abc)^{\frac{1}{3}}+0.75(a+b+c))(a+b)(b+c)(c+a)}+\frac{(c^2+a^2+b^2-ca-ba-cb)(a+b+c)}{(a+b)(b+c)(c+a)}+\frac{9}{4}\left(abc\frac{\frac{a^2+b^2+c^2}{ab+bc+ca}-1}{a^3+b^3+c^3}\right)^2}\geq\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}}$$