49th International Mathematical Olympiad Spain

Solution 1. Denote the four terms by $$A=\frac{(a-b)(a-c)}{a+b+c},B=\frac{(b-c)(b-d)}{b+c+d},C=\frac{(c-d)(c-a)}{c+d+a},D=\frac{(d-a)(d-b)}{d+a+b}.$$ The expression $2A$ splits into two summands as follows, $$2A=A^{\prime }+A^{\prime \prime }\text{ where }{A}^{\prime }=\frac{(a-c{)}^2}{a+b+c},A^{\prime \prime }=\frac{(a-c)(a-2b+c)}{a+b+c}$$ this is easily verified. We analogously represent $2B=B^{\prime }+B^{\prime \prime }$, $2C=C^{\prime }+C^{\prime \prime }$, $2D=D^{\prime }+D^{\prime \prime }$ and examine each of the sums $A^{\prime }+B^{\prime }+C^{\prime }+D^{\prime }$ and $A^{\prime \prime }+B^{\prime \prime }+C^{\prime \prime }+D^{\prime \prime }$ separately. Write $s=a+b+c+d$; the denominators become $s-d$, $s-a$, $s-b$, $s-c$. By the Cauchy-Schwarz inequality, $$\begin{array}{rl}& {\left(\frac{|a-c|}{\sqrt{s-d}}\cdot \sqrt{s-d}+\frac{|b-d|}{\sqrt{s-a}}\cdot \sqrt{s-a}+\frac{|c-a|}{\sqrt{s-b}}\cdot \sqrt{s-b}+\frac{|d-b|}{\sqrt{s-c}}\cdot \sqrt{s-c}\right)}^2\\ & \le \left(\frac{(a-c{)}^2}{s-d}+\frac{(b-d{)}^2}{s-a}+\frac{(c-a{)}^2}{s-b}+\frac{(d-b{)}^2}{s-c}\right)(4s-s)=3s\left(A^{\prime }+B^{\prime }+C^{\prime }+D^{\prime }\right)\end{array}$$ Hence $$\begin{array}{c}{A}^{\prime }+B^{\prime }+C^{\prime }+D^{\prime }\ge \frac{(2|a-c|+2|b-d|{)}^2}{3s}\ge \frac{16\cdot |a-c|\cdot |b-d|}{3s}\end{array}\text{\hspace{1em}(1)}$$ Next we estimate the absolute value of the other sum. We couple $A^{\prime \prime }$ with $C^{\prime \prime }$ to obtain $$\begin{array}{rl}{A}^{\prime \prime }+C^{\prime \prime }& =\frac{(a-c)(a+c-2b)}{s-d}+\frac{(c-a)(c+a-2d)}{s-b}\\ & =\frac{(a-c)(a+c-2b)(s-b)+(c-a)(c+a-2d)(s-d)}{(s-d)(s-b)}\\ & =\frac{(a-c)(-2b(s-b)-b(a+c)+2d(s-d)+d(a+c))}{s(a+c)+bd}\\ & =\frac{3(a-c)(d-b)(a+c)}{M},\text{ with }M=s(a+c)+bd\end{array}$$ Hence by cyclic shift $$B^{\prime \prime }+D^{\prime \prime }=\frac{3(b-d)(a-c)(b+d)}{N},\text{ with }N=s(b+d)+ca$$ Thus $$\begin{array}{c}{A}^{\prime \prime }+B^{\prime \prime }+C^{\prime \prime }+D^{\prime \prime }=3(a-c)(b-d)\left(\frac{b+d}{N}-\frac{a+c}{M}\right)=\frac{3(a-c)(b-d)W}{MN}\end{array}\text{\hspace{1em}(2)}$$ where $$\begin{array}{c}W=(b+d)M-(a+c)N=bd(b+d)-ac(a+c)\end{array}\text{\hspace{1em}(3)}$$ Note that $$\begin{array}{c}MN>ac(a+c)+bd(b+d)s\ge |W|\cdot s.\end{array}\text{\hspace{1em}(4)}$$ Now (2) and (4) yield $$\begin{array}{c}\left|A^{\prime \prime }+B^{\prime \prime }+C^{\prime \prime }+D^{\prime \prime }\right|\le \frac{3\cdot |a-c|\cdot |b-d|}{s}.\end{array}\text{\hspace{1em}(5)}$$ Combined with (1) this results in $$\begin{array}{rl}& 2(A+B+C+D)=\left(A^{\prime }+B^{\prime }+C^{\prime }+D^{\prime }\right)+\left(A^{\prime \prime }+B^{\prime \prime }+C^{\prime \prime }+D^{\prime \prime }\right)\\ & \ge \frac{16\cdot |a-c|\cdot |b-d|}{3s}-\frac{3\cdot |a-c|\cdot |b-d|}{s}=\frac{7\cdot |a-c|\cdot |b-d|}{3(a+b+c+d)}\ge 0\end{array}$$ This is the required inequality. From the last line we see that equality can be achieved only if either $a=c$ or $b=d$. Since we also need equality in (1), this implies that actually $a=c$ and $b=d$ must hold simultaneously, which is obviously also a sufficient condition.

Solution 2. We keep the notations $A$, $B$, $C$, $D$, $s$, and also $M$, $N$, $W$ from the preceding solution; the definitions of $M$, $N$, $W$ and relations (3), (4) in that solution did not depend on the foregoing considerations. Starting from $$2A=\frac{(a-c{)}^2+3(a+c)(a-c)}{a+b+c}-2a+2c,$$ we get $$\begin{array}{rl}& 2(A+C)=(a-c{)}^2\left(\frac{1}{s-d}+\frac{1}{s-b}\right)+3(a+c)(a-c)\left(\frac{1}{s-d}-\frac{1}{s-b}\right)\\ & =(a-c{)}^2\frac{2s-b-d}{M}+3(a+c)(a-c)\cdot \frac{d-b}{M}=\frac{p(a-c{)}^2-3(a+c)(a-c)(b-d)}{M}\end{array}$$ where $p=2s-b-d=s+a+c$. Similarly, writing $q=s+b+d$ we have $$2(B+D)=\frac{q(b-d{)}^2-3(b+d)(b-d)(c-a)}{N};$$ specific grouping of terms in the numerators has its aim. Note that $pq>2s^2$. By adding the fractions expressing $2(A+C)$ and $2(B+D)$, $$2(A+B+C+D)=\frac{p(a-c{)}^2}{M}+\frac{3(a-c)(b-d)W}{MN}+\frac{q(b-d{)}^2}{N}$$ with $W$ defined by (3). Substitution $x=(a-c)/M$, $y=(b-d)/N$ brings the required inequality to the form $$\begin{array}{c}2(A+B+C+D)=Mp{x}^2+3Wxy+Nq{y}^2\ge 0.\end{array}\text{\hspace{1em}(6)}$$ It will be enough to verify that the discriminant $\Delta =9W^2-4MNpq$ of the quadratic trinomial $Mp{t}^2+3Wt+Nq$ is negative; on setting $t=x/y$ one then gets (6). The first inequality in (4) together with $pq>2s^2$ imply $4MNpq>8s^3(ac(a+c)+bd(b+d))$. Since $$(a+c)s^3>(a+c{)}^4\ge 4ac(a+c{)}^2\text{ and likewise }(b+d)s^3>4bd(b+d{)}^2$$ the estimate continues as follows, $$4MNpq>8\left(4(ac{)}^2(a+c{)}^2+4(bd{)}^2(b+d{)}^2\right)>32(bd(b+d)-ac(a+c){)}^2=32W^2\ge 9W^2.$$ Thus indeed $\Delta <0$. The desired inequality (6) hence results. It becomes an equality if and only if $x=y=0$; equivalently, if and only if $a=c$ and simultaneously $b=d$.

Solution 3. $$\begin{array}{c}(a-b)(a-c)(a+b+d)(a+c+d)(b+c+d)=\\ =((a-b)(a+b+d))((a-c)(a+c+d))(b+c+d)=\\ =\left(a^2+ad-b^2-bd\right)\left(a^2+ad-c^2-cd\right)(b+c+d)=\\ =\left(a^4+2a^{3}d-a^2{b}^2-a^{2}bd-a^2{c}^2-a^{2}cd+a^2{d}^2-a{b}^{2}d-ab{d}^2-a{c}^{2}d-ac{d}^2+b^2{c}^2+b^{2}cd+b{c}^{2}d+bc{d}^2\right)(b+c+d)=\\ a^{4}b+a^{4}c+a^{4}d+\left(b^3{c}^2+a^2{d}^3\right)-a^2{c}^3+\left(2a^3{d}^2-b^3{a}^2+c^3{b}^2\right)+\\ +\left(b^{3}cd-c^{3}da-d^{3}ab\right)+\left(2a^{3}bd+c^{3}db-d^{3}ac\right)+\left(2a^{3}cd-b^{3}da+d^{3}bc\right)\\ +\left(-a^2{b}^{2}c+3b^2{c}^{2}d-2a{c}^2{d}^2\right)+\left(-2a^2{b}^{2}d+2b{c}^2{d}^2\right)+\left(-a^{2}b{c}^2-2a^2{c}^{2}d-2a{b}^2{d}^2+2b^{2}c{d}^2\right)+\\ +\left(-2a^{2}bcd-a{b}^{2}cd-ab{c}^{2}d-2abc{d}^2\right)\end{array}$$ Introducing the notation $S_{xyzw}={\sum }_{cyc}{a}^x{b}^y{c}^z{d}^w$, one can write $$\begin{array}{c}\sum _{cyc}(a-b)(a-c)(a+b+d)(a+c+d)(b+c+d)=\\ =S_{4100}+S_{4010}+S_{4001}+2S_{3200}-S_{3020}+2S_{3002}-S_{3110}+2S_{3101}+2S_{3011}-3S_{2120}-6S_{2111}=\\ +\left(S_{4100}+S_{4001}+\frac{1}{2}{S}_{3110}+\frac{1}{2}{S}_{3011}-3S_{2120}\right)+\\ +\left(S_{4010}-S_{3020}-\frac{3}{2}{S}_{3110}+\frac{3}{2}{S}_{3011}+\frac{9}{16}{S}_{2210}+\frac{9}{16}{S}_{2201}-\frac{9}{8}{S}_{2111}\right)+\\ +\frac{9}{16}\left(S_{3200}-S_{2210}-S_{2201}+S_{3002}\right)+\frac{23}{16}\left(S_{3200}-2S_{3101}+S_{3002}\right)+\frac{39}{8}\left(S_{3101}-S_{2111}\right),\end{array}$$ where the expressions $$\begin{array}{c}{S}_{4100}+S_{4001}+\frac{1}{2}{S}_{3110}+\frac{1}{2}{S}_{3011}-3S_{2120}=\sum _{cyc}\left(a^{4}b+b{c}^4+\frac{1}{2}{a}^{3}bc+\frac{1}{2}ab{c}^3-3a^{2}b{c}^2\right),\\ S_{4010}-S_{3020}-\frac{3}{2}{S}_{3110}+\frac{3}{2}{S}_{3011}+\frac{9}{16}{S}_{2210}+\frac{9}{16}{S}_{2201}-\frac{9}{8}{S}_{2111}=\sum _{cyc}{a}^{2}c{\left(a-c-\frac{3}{4}b+\frac{3}{4}d\right)}^2,\\ S_{3200}-S_{2210}-S_{2201}+S_{3002}=\sum _{cyc}{b}^2\left(a^3-a^{2}c-a{c}^2+c^3\right)=\sum _{cyc}{b}^2(a+c)(a-c{)}^2,\\ S_{3200}-2S_{3101}+S_{3002}=\sum _{cyc}{a}^3(b-d{)}^2\text{ and }{S}_{3101}-S_{2111}=\frac{1}{3}\sum _{cyc}bd\left(2a^3+c^3-3a^{2}c\right)\end{array}$$ are all nonnegative.