Without loss of generality, we can assume that $ a\geq c$, $ b\geq d$. Then $ \frac{(a-b)(a-c)}{a+b+c}+\frac{(b-c)(b-d)}{b+c+d}+\frac{(c-d)(c-a)}{c+d+a}+\frac{(d-a)(d-b)}{d+a+b}$ $ =\frac{(a-c)^2}{a+c+d}+\frac{(b-d)^2}{a+b+d}-\frac{(a-b)(a-c)(b-d)(c^2+d^2+2ac+ad+bc+2bd}{(a+b+c)(a+b+d)(a+c+d)(b+c+d)}$. If $ a\leq b$, then $ (a-b)(a-c)(b-d)\leq 0$, and if $ a\geq c$, $ b\geq c$, then $ (b-c)(b-d)(c-a)\leq 0$. Therefore, in these cases, we proved the inequality. We only need prove the inequality when $ a\geq c\geq b\geq d$. Let $ x=a-c$, $ y=c-b$, $ z=b-d$. Then the inequality is equal to $ \frac{x^2}{a+b+c}+\frac{z^2}{a+b+d}+xyz\left(\frac{1}{(a+c+d)(a+b+d)}-\frac{1}{(a+b+c)(a+c+d)}-\frac{1}{(a+b+d)(b+c+d)}\right)$ $ \geq 0$. We know that $ \frac{1}{a+b+c}\geq \frac{1}{b+c+d}\cdot \frac{y+2z}{x+2y+3z}\geq \frac{1}{b+c+d}\cdot \frac{y}{x+2y+3z}$. Therefore, $ \frac{x^2}{a+b+c}+\frac{z^2}{a+b+d}\geq \frac{x^2(x+y+2z)+z^2(x+2y+3z)}{(a+b+c)(a+b+d)}$ $ \geq\frac{(x^2+z^2)(x+2y+3z)}{2(a+b+c)(a+b+d)}\geq \frac{xyz}{(a+b+d)(b+c+d)}$. Also, because $ (a+b+c)(a+c+d)\geq (a+c+d)(a+b+d)$, $ xyz\left(\frac{1}{(a+c+d)(a+b+d)}-\frac{1}{(a+b+c)(a+c+d)}\right)\geq 0$. Therefore, we proved the inequality when $ a\geq c\geq b\geq d$. It is easy to see that the equality holds if and only if $ a=c$ and $ b=d$.

I will write just the main steps because I'm really tired since it's 3am WLOG $ a = max(a,b,c,d)$ note that the inequality is equivalent to : $ \frac {(a - d)(b - c)(a + 2d)}{(d + a + b)(c + a + d)} + \frac {(a - d)(b - c)(2b + c)}{(a + b + c)(b + c + d)} + \frac {(c - d)^2}{c + d + a} + \frac {(a - b)^2}{a + b + c}$ hence it remain to prove the inequality in the case $ c \geq b$ if $ a\geq d \geq c \geq b$ we have : $ LHS \geq \frac {(a - b)(a - c)}{a + b + c} + \frac {(d - a)(d - b)}{d + a + b} \geq (d - b) ( \frac {a - c}{a + b + c} + \frac {d - a}{b + a + d} ) \geq \frac { (d - b)(d - c)}{b + a + d} \geq 0$ if $ a\geq c \geq d \geq b$ case 1 : $ a - c \leq d - b$ hence : $ a - d \leq c - b$ thus $ LHS \geq (a - c)( \frac {a - b}{a + b + c} - \frac { c - d}{c + d + a} ) + \frac { (a - c)(d - b)(c - b)}{(b + c + d)(d + a + b)} \geq 0$ case 2 : $ a - c \geq d - b$ then the inequality is equivalent to : $ (a - c)( \frac { a - b}{a + b + c} - \frac {c - d}{c + d + a} ) + (d - b)(\frac {d - a}{d + a + b} + \frac {c - b}{b + c + d} ) \geq 0$ thus $ LHS \geq (d - b)( \frac { a - b}{a + b + c} - \frac {c - d}{c + d + a} + \frac {d - a}{d + a + b} + \frac {c - b}{b + c + d} ) \geq 0$ easy to prove !! if $ a\geq c \geq b \geq d$ note that the inequality is equivalent to : $ (a - c)( \frac { a - b}{a + b + c} - \frac {c - d}{c + d + a} ) + (b - d) (\frac {a - d}{d + a + b} + \frac {b - c}{b + c + d} ) \geq 0$ now $ \frac {a - d}{d + a + b} + \frac {b - c}{b + c + d} \geq 0$ just let : $ a = d + x + y + z ,c = d + x + y....$ hence we have 2 cases : if $ b - d \geq a - c$ as the previous method the inequality is easy to prove if $ a - c \geq b - d$ note that : $ \frac { a - b}{a + b + c} - \frac {c - d}{c + d + a} \geq 0$ is not true so we will use the trivial identity : $ \frac { (a - c)(b - d)}{a + b + c} + \frac { (b - d)(c - a)}{a + b + c} \geq 0$ hence the inequality is equivalent to : $ (a - c)( \frac { a - b}{a + b + c} - \frac {c - d}{c + d + a} + \frac {b - d}{a + b + c} ) + (b - d)(\frac {a - d}{d + a + b} + \frac {b - c}{b + c + d} + \frac {c - a}{a + b + c} ) \geq 0$ it is easy to prove that :$ \frac{ a-b}{a+b+c} -\frac{c-d}{c+d+a} + \frac{b-d}{a+b+c} \geq 0$with the condition : $ a - c \geq b - d$ thus : $ LHS \geq (b - d)( \frac {a - d}{b + c + a} - \frac {c - d}{c + d + a} + \frac {a - d}{d + a + b} + \frac {b - c}{b + c + d} + \frac {c - a}{a + b + c}) \geq 0$ the last inequality can be handeled with easy manipulations , hence the inequality is proved omg what an ugly proof

April wrote: Prove that for any four positive real numbers $ a$, $ b$, $ c$, $ d$ the inequality \[ \frac {(a - b)(a - c)}{a + b + c} + \frac {(b - c)(b - d)}{b + c + d} + \frac {(c - d)(c - a)}{c + d + a} + \frac {(d - a)(d - b)}{d + a + b}\ge 0\] holds. Determine all cases of equality. Author: Darij Grinberg (Problem Proposal), Christian Reiher (Solution), Germany Dear friend, please take a look at: http://mathvn.org/forum/viewthread.php?thread_id=1254 (problem O23, page 23).

My solution(very sorry if repeated): Firstly it is equivalent to $ \frac{(a-c)^2}{c+d+a}+\frac{(b-d)^2}{d+a+b}+(a-c)(b-d)(\frac{2b+d}{(a+b+d)(b+c+d)}-\frac{2a+c}{(a+b+c)(c+d+a)}\ge 0$ We only need to prove:$ (\frac{2b+d}{(a+b+d)(b+c+d)}-\frac{2a+c}{(a+b+c)(c+d+a)})^2\ge \frac{4}{(a+c+d)(b+d+a)}$ We are sufficient to show both $ (\frac{2b+d}{(a+b+d)(b+c+d)})^2$ and $ (\frac{2a+c}{(a+b+c)(c+d+a)})^2$ are not bigger than $ \frac{4}{(a+c+d) (b+d+a)}$ They are almost trivial if we notice $ (2b+d)^2 \le 4(b+d)^2$ and some easy calculation

can_hang2007 wrote: April wrote: Prove that for any four positive real numbers $ a$, $ b$, $ c$, Author: Darij Grinberg (Problem Proposal), Christian Reiher (Solution), Germany Dear friend, please take a look at: http://mathvn.org/forum/viewthread.php?thread_id=1254 (problem O23, page 23). Dear can_hang2007 I canot open the file .pdf , please can you put here

My solution is more computational than the ones provided above, but in my opinion it is also a bit more motivated (though perhaps this could probably just be attributed to my lack of experience with inequalities.)

math154 has pointed out that there is an error in this solution. $a^4 + 2ac^3 + c^4 - a^3 c - 3a^2 c^2 \geq 0$ is false (and is completely irrelevant to my lemma .) Here is a revised proof, hopefully with no mistakes this time:

A7 in 2008 seems to be easy. From $ a,b,c,d \in\mathbb{R}^+ $, we yield \[ \frac{(a-b)(a-c)}{a+b+c}\geq\frac{(a-b)(a-c)}{a+b+c+d}=\frac{a^2-ab-ac+bc}{a+b+c+d} \] Similarly, we got \[ \frac{(b-c)(b-d)}{b+c+d}\geq\frac{b^2-bc-bd+cd}{a+b+c+d} \] \[ \frac{(c-d)(c-a)}{c+d+a}\geq\frac{c^2-cd-ca+ad}{a+b+c+d} \] \[ \frac{(d-a)(d-b)}{d+a+b}\geq\frac{d^2-ad-bd+ab}{a+b+c+d} \] Sum all of the above inequalities together, we obtain: \[ \frac{(a-b)(a-c)}{a+b+c}+\frac{(b-c)(b-d)}{b+c+d}+\frac{(c-d)(c-a)}{c+d+a}+\frac{(d-a)(d-b)}{d+a+b}\geq\frac{a^2-ab-ac+bc+b^2-bc-bc+cd+c^2-cd-ca+ad+d^2-da-db+ab}{a+b+c+d}=\frac{a^2+b^2+c^2+d^2}{a+b+c+d}\geq 0 \] Equality occurs iff $ a=c $ or $ b=d $

quangminhltv99 wrote: A7 in 2008 seems to be easy. From $ a,b,c,d \in\mathbb{R}^+ $, we yield \[ \frac{(a-b)(a-c)}{a+b+c}\geq\frac{(a-b)(a-c)}{a+b+c+d}=\frac{a^2-ab-ac+bc}{a+b+c+d} \] Well, no. That holds only if the fractions are nonnegative. For example if $c>a>b$, then it doesn't hold.

hxy09 wrote: My solution(very sorry if repeated): Firstly it is equivalent to $ \frac{(a-c)^2}{c+d+a}+\frac{(b-d)^2}{d+a+b}+(a-c)(b-d)(\frac{2b+d}{(a+b+d)(b+c+d)}-\frac{2a+c}{(a+b+c)(c+d+a)}\ge 0$ What would the motivation for splitting the expression in this way be? Thanks!

Just a query; after clearing the denominators, isn't it pretty much do-able by Lagrange Multipliers(of course, after normalizing)

April wrote: Prove that for any four positive real numbers $ a$, $ b$, $ c$, $ d$ the inequality \[ \frac {(a - b)(a - c)}{a + b + c} + \frac {(b - c)(b - d)}{b + c + d} + \frac {(c - d)(c - a)}{c + d + a} + \frac {(d - a)(d - b)}{d + a + b}\ge 0\]holds. Determine all cases of equality. Author: Darij Grinberg (Problem Proposal), Christian Reiher (Solution), Germany We have \[LHS=\frac{1}{660}\frac{(4a^2b+297a^2c+20ab^2+490abc+297ac^2+490acd+144ad^2+144b^2c+77b^2d+77bd^2+4c^2d+20cd^2)(a-2b+2d-c)^2}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}\]\[+\frac{1}{660}\frac{(144a^2b+77a^2c+20a^2d+490abd+77ac^2+4ad^2+4b^2c+297b^2d+20bc^2+490bcd+297bd^2+144c^2d)(2a+b-2c-d)^2}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}\]\[+\frac{1}{132}\frac{(11a^2c+116a^2d+232abd+11ac^2+16ad^2+16b^2c+116bc^2+232bcd)(a-b-c+d)^2}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}\]\[+\frac{1}{132}\frac{(16a^2b+116ab^2+232abc+232acd+11b^2d+11bd^2+16c^2d+116cd^2)(a+b-c-d)^2}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}\]\[\ge{0}\]

well, hxy09... hxy09 wrote: We only need to prove$ (\frac{2b+d}{(a+b+d)(b+c+d)}-\frac{2a+c}{(a+b+c)(c+d+a)})^2\ge \frac{4}{(a+c+d)(b+d+a)}$ But, this does not satisfy when $a=b=c=d$!! .....isn't it!?

Note. The term ``LHS'' will always refer to the expression $\frac {(a - b)(a - c)}{a + b + c} + \frac {(b - c)(b - d)}{b + c + d} + \frac {(c - d)(c - a)}{c + d + a} + \frac {(d - a)(d - b)}{d + a + b}$. We will rewrite the LHS as \[ (a-c)(\frac{a-b}{a+b+c} + \frac{d-c}{c+d+a}) + (b-d)(\frac{b-c}{b+c+d} + \frac{a-d}{d+a+b}). \]Let us focus on the first term $(a-c)(\frac{a-b}{a+b+c} + \frac{d-c}{c+d+a})$. We will try to express this in terms of $(a-c)$ and $(b-d)$. Thus we write \[(a-c)(\frac{a-b}{a+b+c} + \frac{d-c}{c+d+a}) = (a-c)(\frac{(ac+ad+a^2-bc-bd-ba) + (da+db+dc-ca-cb-c^2)}{(a+b+c)(c+d+a)}) \]\[ = (a-c)(\frac{2ad - 2bc - ab + cd + a^2 - c^2}{(a+b+c)(c+d+a)}) \]\[ = (a-c)(\frac{(a-c)(a+c+\frac{1}{2}b+\frac{1}{2}d) + (b-d)(-\frac{3}{2}a-\frac{3}{2}c)}{(a+b+c)(c+d+a)}) \]\[ = (a-c)^2\frac{a+c+\frac{1}{2}b+\frac{1}{2}d}{(a+b+c)(c+d+a)} + (a-c)(b-d)\frac{-\frac{3}{2}a-\frac{3}{2}c}{(a+b+c)(c+d+a)}\] Similarly, the second term $(b-d)(\frac{b-c}{b+c+d} + \frac{a-d}{d+a+b})$ equals \[ (b-d)^2\frac{b+d+\frac{1}{2}a+\frac{1}{2}c}{(b+c+d)(d+a+b)} + (a-c)(b-d)\frac{\frac{3}{2}b+\frac{3}{2}d}{(b+c+d)(d+a+b)} \] Thus, the entire LHS equals \[(a-c)^2X + (b-d)^2Z + (a-c)(b-d)Y,\]where $X = \frac{a+c+\frac{1}{2}b+\frac{1}{2}d}{(a+b+c)(c+d+a)}$, $Y = \frac{3}{2}(\frac{b+d}{(b+c+d)(d+a+b)} - \frac{a+c}{(a+b+c)(c+d+a)})$, and $Z = \frac{b+d+\frac{1}{2}a+\frac{1}{2}c}{(b+c+d)(d+a+b)}$. Let's simplify $Y$ a little bit. Miraculously, we can write \[Y = \frac{3}{2}(\frac{(b+d)(a+b+c)(c+d+a) - (a+c)(b+c+d)(d+a+b)}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)})\]\[ = \frac{3}{2}(\frac{(b+d)(bd+(b+d)(a+c)+(a+c)^2) - (a+c)(ac+(a+c)(b+d) + (b+d)^2)}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)})\]\[ = \frac{3}{2}(\frac{bd(b+d)-ac(a+c)}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}).\]Note how most of the terms in the numerator cancel. Let $u = a-c$, $v = b-d$. Then the LHS is \[u^2X + v^2Z + uvY,\]which is quadratic in $u,v$. I claim that $Y^2 - 4XZ < 0$, or equivalently: Lemma. $Y^2 < 4XZ$ Proof. To prove this, we first note that \[4XZ = \frac{(2a+2c+b+d)(2b+2d+a+c)}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}\]and \[Y^2 = \frac{9}{4}(\frac{(bd(b+d)-ac(a+c))^2}{((a+b+c)(b+c+d)(c+d+a)(d+a+b))^2}).\]Thus $Y^2 < 4XZ$ is equivalent to \[\frac{9}{4}(bd(b+d)-ac(a+c))^2 < (2a+2c+b+d)(2b+2d+a+c)(a+b+c)(b+c+d)(c+d+a)(d+a+b)\] Note that, if $x,y > 0$, $(x-y)^2 < x^2$ if $x \ge y$ and $(x-y)^2 < y^2$ if $y \ge x$. Let's let $x = ac(a+c)$, $y = bd(b+d)$. Then we can assume WLOG that $x \ge y$. We have \[\frac{9}{4}(bd(b+d)-ac(a+c))^2 < \frac{9}{4}(ac(a+c))^2\] But we also have \[(2a+2c+b+d)(2b+2d+a+c)(a+b+c)(b+c+d)(c+d+a)(d+a+b) > (2a+2c)(a+c)(a+c)(c)(a+c)(a) = 2ac(a+c)^4\] Thus, if we can show that \[2ac(a+c)^4 > \frac{9}{4}(ac(a+c))^2,\]we would be done with the proof of this Lemma. But this is equivalent to \[(a+c)^2 > \frac{9}{8}ac,\]which of course is true. $\blacksquare$ Now that we have $Y^2 - 4XZ < 0$, let's show that the LHS, which is $u^2X + v^2Z + uvY$, is always nonnegative. Note that $X,Z > 0$. First suppose that $u = v = 0$; then obviously the LHS equals $0$, which is nonnegative. Now suppose that $u = 0$ but $v \not= 0$. Then the LHS is $v^2Z$, which is positive. Similarly, if $u \not= 0 $ and $v = 0$, then the LHS is $u^2X$, which is positive. Now suppose that $u,v \not= 0 $. We can write the LHS as \[v^2(X(u/v)^2 + Y(u/v) + Z)\]Since $Y^2 - 4XZ < 0$, we know that $Xr^2 + Yr + Z > 0$ for all real numbers $r$. This means that $X(u/v)^2 + Y(u/v) + Z > 0$. Since $v^2 > 0$, we know that the LHS is positive. Thus the LHS is always nonnegative, with equality occurring if and only if $u = v = 0$, which is equivalent to $a=c$ and $b=d$ holding simultaneously. $\blacksquare$

Claim: [The main rewrite] The given inequality may be rewritten as \begin{align*} (a-c)^2 & \left( \frac{1}{a+b+c} + \frac{1}{c+d+a} \right) + (b-d)^2 \left( \frac{1}{b+c+d} + \frac{1}{d+a+b} \right) \\ &\ge 3(a-c)(b-d) \left[ \frac{a+c}{(a+b+c)(c+d+a)} - \frac{b+d}{(b+c+d)(d+a+b)} \right]. \end{align*}

In light of this, taking the absolute value of right-hand side, it suffices to prove \begin{align*} (a-c)^2 & \left( \frac{1}{a+b+c} + \frac{1}{c+d+a} \right) + (b-d)^2 \left( \frac{1}{b+c+d} + \frac{1}{d+a+b} \right) \\ &\ge 3 |a-c| \cdot |b-d| \cdot \max \left\{ \frac{a+c}{(a+b+c)(c+d+a)}, \frac{b+d}{(b+c+d)(d+a+b)} \right\}. \end{align*}We show only the first one (other one ditto). In fact, a crude AM-GM suffices: Claim: We have \[ 4\sqrt[4]{\prod_{\text{cyc}} \frac{1}{a+b+c}} > \frac{3(a+c)}{(a+b+c)(c+d+a)}. \]

And thus we are done. Equality holds iff $|a-c| = |b-d| = 0$, which is $a=c$ and $b=d$.

\[ \frac{(a-b)(a-c)}{a+b+c}+\frac{(b-c)(b-d)}{b+c+d}+\frac{(c-d)(c-a)}{c+d+a}+\frac{(d-a)(d-b)}{d+a+b}\geq 0 \]\[ \frac{2(a-b)(a-c)}{a+b+c}+\frac{2(b-c)(b-d)}{b+c+d}+\frac{2(c-d)(c-a)}{c+d+a}+\frac{2(d-a)(d-b)}{d+a+b}\geq 0 \]\[(\frac{2(a-b)(a-c)}{a+b+c}+\frac{(2b-c-a)(a-c)}{a+b+c})+(\frac{2(b-c)(b-d)}{b+c+d}+\frac{(2c-d-b)(b-d)}{b+c+d})+(\frac{2(c-d)(c-a)}{c+d+a}+\frac{(2d-a-c)(c-a)}{c+d+a})+(\frac{2(d-a)(d-b)}{d+a+b}+\frac{(2a-b-d)(d-b)}{d+a+b})\geq \]\[\geq\frac{(2b-c-a)(a-c)}{a+b+c}+\frac{(2c-d-b)(b-d)}{b+c+d}+\frac{(2d-a-c)(c-a)}{c+d+a}+\frac{(2a-b-d)(d-b)}{d+a+b} \]\[\frac{(a-c)^2}{a+b+c}+\frac{(b-d)^2}{b+c+d}+\frac{(c-a)^2}{c+d+a}+\frac{(d-b)^2}{d+a+b}\geq (a-c)(\frac{2b-c-a}{a+b+c}-\frac{2d-a-c}{c+d+a})+(b-d)(\frac{2c-d-b}{b+c+d}-\frac{2a-b-d}{d+a+b})\]Transform the right side of the inequality: \[(a-c)(\frac{2b-c-a}{a+b+c}-\frac{2d-a-c}{c+d+a})+(b-d)(\frac{2c-d-b}{b+c+d}-\frac{2a-b-d}{d+a+b})=(a-c)(\frac{3b}{a+b+c}-\frac{3d}{c+d+a})+(b-d)(\frac{3c}{b+c+d}-\frac{3a}{d+a+b})= \]\[=\frac{3(a-c)(b(c+d+a)-d(a+b+c))}{(a+b+c)(c+d+a)}+\frac{3(b-d)(c(d+a+b)-a(b+c+d))}{(b+c+d)(d+a+b)}=\frac{3(a-c)(b-d)(a+c)}{(a+b+c)(c+d+a)}+\frac{3(b-d)(c-a)(b+d)}{(b+c+d)(d+a+b)}=\]\[=\frac{3(a-c)(b-d)((a+c)(b+c+d)(d+a+b)-(b+d)(a+b+c)(c+d+a))}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}=\]\[=\frac{3(a-c)(b-d)((a+c)(b+d)^2+(a+c)^2(b+d)+ac(a+c)-(b+d)(a+c)^2-(b+d)^2(a+c)-bd(b+d))}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}=\]\[=\frac{3(a-c)(b-d)(ac(a+c)-bd(b+d))}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)} \]We need to prove that: \[\frac{(a-c)^2}{a+b+c}+\frac{(b-d)^2}{b+c+d}+\frac{(c-a)^2}{c+d+a}+\frac{(d-b)^2}{d+a+b}\geq \frac{3(a-c)(b-d)(ac(a+c)-bd(b+d))}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)} \]We write a set of such inequalities: \[a*c*(a+c)<(a+b+c)*(b+c+d)*(c+d+a)=\frac{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}{d+a+b} \]\[a*c*(a+c)<(d+a+b)*(b+c+d)*(a+b+c)=\frac{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}{c+d+a} \]\[a*c*(a+c)<(d+a+b)*(a+b+c)*(c+d+a)=\frac{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}{b+c+d} \]\[a*c*(a+c)<(d+a+b)*(b+c+d)*(c+d+a)=\frac{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}{a+b+c} \]We get: \[ac(a+c)<\frac{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}{max\{a+b+c;b+c+d;c+d+a;d+a+b\}}\]Similarly: \[b*d*(b+d)<(a+b+c)*(c+d+a)*(b+c+d)=\frac{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}{d+a+b} \]\[b*d*(b+d)<(a+b+c)*(d+a+b)*(b+c+d)=\frac{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}{c+d+a} \]\[b*d*(b+d)<(a+b+c)*(c+d+a)*(d+a+b)=\frac{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}{b+c+d} \]\[b*d*(b+d)<(d+a+b)*(c+d+a)*(b+c+d)=\frac{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}{a+b+c} \]We get: \[bd(b+d)<\frac{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}{max\{a+b+c;b+c+d;c+d+a;d+a+b\}}\]Combine 2 inequalities obtained: \[max\{ac(a+c);bd(b+d)\}<\frac{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}{max\{a+b+c;b+c+d;c+d+a;d+a+b\}} \]We still need the following inequality, which we prove right away: \[2(a-c)^2+2(b-d)^2\geq 3*|a-c|*|b-d| \]\[\frac{1}{2}(a-c)^2+\frac{3}{2}(|a-c|-|b-d|)^2+\frac{1}{2}(b-d)^2\geq 0 \]Now we can prove the main inequality: \[ \frac{3(a-c)(b-d)(ac(a+c)-bd(b+d))}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}\leq \frac{3*|a-c|*|b-d|*|ac(a+c)-bd(b+d)|}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}\leq \frac{3*|a-c|*|b-d|*(max\{ac(a+c);bd(b+d)\}}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)} \leq \]\[\leq \frac{3*|a-c|*|b-d|}{max\{a+b+c;b+c+d;c+d+a;d+a+b\}}\leq \frac{2(a-c)^2+2(b-d)^2}{max\{a+b+c;b+c+d;c+d+a;d+a+b\}} \leq \frac{(a-c)^2}{a+b+c}+\frac{(b-d)^2}{b+c+d}+\frac{(c-a)^2}{c+d+a}+\frac{(d-b)^2}{d+a+b} \]Equality condition: a=c;b=d