Subfield: Algebra


Question: Prove that for any four positive real numbers $a, b, c, d$ the inequality

$$
\frac{(a-b)(a-c)}{a+b+c}+\frac{(b-c)(b-d)}{b+c+d}+\frac{(c-d)(c-a)}{c+d+a}+\frac{(d-a)(d-b)}{d+a+b} \geq 0
$$

holds. Determine all cases of equality.

Solution:
Denote the four terms by

$$
A=\frac{(a-b)(a-c)}{a+b+c}, \quad B=\frac{(b-c)(b-d)}{b+c+d}, \quad C=\frac{(c-d)(c-a)}{c+d+a}, \quad D=\frac{(d-a)(d-b)}{d+a+b}
$$

The expression $2 A$ splits into two summands as follows,

$$
2 A=A^{\prime}+A^{\prime \prime} \quad \text { where } \quad A^{\prime}=\frac{(a-c)^{2}}{a+b+c}, \quad A^{\prime \prime}=\frac{(a-c)(a-2 b+c)}{a+b+c}
$$

this is easily verified. We analogously represent $2 B=B^{\prime}+B^{\prime \prime}, 2 C=C^{\prime}+C^{\prime \prime}, 2 B=D^{\prime}+D^{\prime \prime}$ 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 CauchySchwarz inequality,

$$
\begin{aligned}
& \left(\frac{|a-c|}{\sqrt{s-d}} \cdot \sqrt{s-d}+\frac{|b-d|}{\sqrt{s-a}} \cdot \sqrt{s-a}+\frac{|c-a|}{\sqrt{s-b}} \cdot \sqrt{s-b}+\frac{|d-b|}{\sqrt{s-c}} \cdot \sqrt{s-c}\right)^{2} \\
& \quad \leq\left(\frac{(a-c)^{2}}{s-d}+\frac{(b-d)^{2}}{s-a}+\frac{(c-a)^{2}}{s-b}+\frac{(d-b)^{2}}{s-c}\right)(4 s-s)=3 s\left(A^{\prime}+B^{\prime}+C^{\prime}+D^{\prime}\right) .
\end{aligned}
$$

Hence

$$
A^{\prime}+B^{\prime}+C^{\prime}+D^{\prime} \geq \frac{(2|a-c|+2|b-d|)^{2}}{3 s} \geq \frac{16 \cdot|a-c| \cdot|b-d|}{3 s}\tag{1}
$$

Next we estimate the absolute value of the other sum. We couple $A^{\prime \prime}$ with $C^{\prime \prime}$ to obtain

$$
\begin{aligned}
A^{\prime \prime}+C^{\prime \prime} & =\frac{(a-c)(a+c-2 b)}{s-d}+\frac{(c-a)(c+a-2 d)}{s-b} \\
& =\frac{(a-c)(a+c-2 b)(s-b)+(c-a)(c+a-2 d)(s-d)}{(s-d)(s-b)} \\
& =\frac{(a-c)(-2 b(s-b)-b(a+c)+2 d(s-d)+d(a+c))}{s(a+c)+b d} \\
& =\frac{3(a-c)(d-b)(a+c)}{M}, \quad \text { with } \quad M=s(a+c)+b d .
\end{aligned}
$$

Hence by cyclic shift

$$
B^{\prime \prime}+D^{\prime \prime}=\frac{3(b-d)(a-c)(b+d)}{N}, \quad \text { with } \quad N=s(b+d)+c a
$$

Thus

$$
A^{\prime \prime}+B^{\prime \prime}+C^{\prime \prime}+D^{\prime \prime}=3(a-c)(b-d)\left(\frac{b+d}{N}-\frac{a+c}{M}\right)=\frac{3(a-c)(b-d) W}{M N}\tag{2}
$$

where

$$
W=(b+d) M-(a+c) N=b d(b+d)-a c(a+c)\tag{3}
$$



Note that

$$
M N>(a c(a+c)+b d(b+d)) s \geq|W| \cdot s\tag{4}
$$

Now (2) and (4) yield

$$
\left|A^{\prime \prime}+B^{\prime \prime}+C^{\prime \prime}+D^{\prime \prime}\right| \leq \frac{3 \cdot|a-c| \cdot|b-d|}{s}\tag{5}
$$

Combined with (1) this results in

$$
\begin{aligned}
& 2(A+B+C+D)=\left(A^{\prime}+B^{\prime}+C^{\prime}+D^{\prime}\right)+\left(A^{\prime \prime}+B^{\prime \prime}+C^{\prime \prime}+D^{\prime \prime}\right) \\
& \quad \geq \frac{16 \cdot|a-c| \cdot|b-d|}{3 s}-\frac{3 \cdot|a-c| \cdot|b-d|}{s}=\frac{7 \cdot|a-c| \cdot|b-d|}{3(a+b+c+d)} \geq 0
\end{aligned}
$$

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.
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

$$
2 A=\frac{(a-c)^{2}+3(a+c)(a-c)}{a+b+c}-2 a+2 c
$$

we get

$$
\begin{aligned}
2(A & +C)=(a-c)^{2}\left(\frac{1}{s-d}+\frac{1}{s-b}\right)+3(a+c)(a-c)\left(\frac{1}{s-d}-\frac{1}{s-b}\right) \\
& =(a-c)^{2} \frac{2 s-b-d}{M}+3(a+c)(a-c) \cdot \frac{d-b}{M}=\frac{p(a-c)^{2}-3(a+c)(a-c)(b-d)}{M}
\end{aligned}
$$

where $p=2 s-b-d=s+a+c$. 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 $p q>2 s^{2}$. By adding the fractions expressing $2(A+C)$ and $2(B+D)$,

$$
2(A+B+C+D)=\frac{p(a-c)^{2}}{M}+\frac{3(a-c)(b-d) W}{M N}+\frac{q(b-d)^{2}}{N}
$$

with $W$ defined by (3).

Substitution $x=(a-c) / M, y=(b-d) / N$ brings the required inequality to the form

$$
2(A+B+C+D)=M p x^{2}+3 W x y+N q y^{2} \geq 0\tag{6}
$$

It will be enough to verify that the discriminant $\Delta=9 W^{2}-4 M N p q$ of the quadratic trinomial $M p t^{2}+3 W t+N q$ is negative; on setting $t=x / y$ one then gets (6). The first inequality in (4) together with $p q>2 s^{2}$ imply $4 M N p q>8 s^{3}(a c(a+c)+b d(b+d))$. Since

$$
(a+c) s^{3}>(a+c)^{4} \geq 4 a c(a+c)^{2} \quad \text { and likewise } \quad(b+d) s^{3}>4 b d(b+d)^{2}
$$

the estimate continues as follows,

$$
4 M N p q>8\left(4(a c)^{2}(a+c)^{2}+4(b d)^{2}(b+d)^{2}\right)>32(b d(b+d)-a c(a+c))^{2}=32 W^{2} \geq 9 W^{2}
$$

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 $b=d$.
$$
\begin{gathered}
(a-b)(a-c)(a+b+d)(a+c+d)(b+c+d)= \\
=((a-b)(a+b+d))((a-c)(a+c+d))(b+c+d)= \\
=\left(a^{2}+a d-b^{2}-b d\right)\left(a^{2}+a d-c^{2}-c d\right)(b+c+d)= \\
=\left(a^{4}+2 a^{3} d-a^{2} b^{2}-a^{2} b d-a^{2} c^{2}-a^{2} c d+a^{2} d^{2}-a b^{2} d-a b d^{2}-a c^{2} d-a c d^{2}+b^{2} c^{2}+b^{2} c d+b c^{2} d+b c d^{2}\right)(b+c+d)= \\
=a^{4} b+a^{4} c+a^{4} d+\left(b^{3} c^{2}+a^{2} d^{3}\right)-a^{2} c^{3}+\left(2 a^{3} d^{2}-b^{3} a^{2}+c^{3} b^{2}\right)+ \\
+\left(b^{3} c d-c^{3} d a-d^{3} a b\right)+\left(2 a^{3} b d+c^{3} d b-d^{3} a c\right)+\left(2 a^{3} c d-b^{3} d a+d^{3} b c\right) \\
+\left(-a^{2} b^{2} c+3 b^{2} c^{2} d-2 a c^{2} d^{2}\right)+\left(-2 a^{2} b^{2} d+2 b c^{2} d^{2}\right)+\left(-a^{2} b c^{2}-2 a^{2} c^{2} d-2 a b^{2} d^{2}+2 b^{2} c d^{2}\right)+ \\
+\left(-2 a^{2} b c d-a b^{2} c d-a b c^{2} d-2 a b c d^{2}\right)
\end{gathered}
$$

Introducing the notation $S_{x y z w}=\sum_{c y c} a^{x} b^{y} c^{z} d^{w}$, one can write

$$
\begin{gathered}
\sum_{c y c}(a-b)(a-c)(a+b+d)(a+c+d)(b+c+d)= \\
=S_{4100}+S_{4010}+S_{4001}+2 S_{3200}-S_{3020}+2 S_{3002}-S_{3110}+2 S_{3101}+2 S_{3011}-3 S_{2120}-6 S_{2111}= \\
+\left(S_{4100}+S_{4001}+\frac{1}{2} S_{3110}+\frac{1}{2} S_{3011}-3 S_{2120}\right)+ \\
+\left(S_{4010}-S_{3020}-\frac{3}{2} S_{3110}+\frac{3}{2} S_{3011}+\frac{9}{16} S_{2210}+\frac{9}{16} S_{2201}-\frac{9}{8} S_{2111}\right)+ \\
+\frac{9}{16}\left(S_{3200}-S_{2210}-S_{2201}+S_{3002}\right)+\frac{23}{16}\left(S_{3200}-2 S_{3101}+S_{3002}\right)+\frac{39}{8}\left(S_{3101}-S_{2111}\right),
\end{gathered}
$$

where the expressions

$$
\begin{gathered}
S_{4100}+S_{4001}+\frac{1}{2} S_{3110}+\frac{1}{2} S_{3011}-3 S_{2120}=\sum_{c y c}\left(a^{4} b+b c^{4}+\frac{1}{2} a^{3} b c+\frac{1}{2} a b c^{3}-3 a^{2} b c^{2}\right), \\
S_{4010}-S_{3020}-\frac{3}{2} S_{3110}+\frac{3}{2} S_{3011}+\frac{9}{16} S_{2210}+\frac{9}{16} S_{2201}-\frac{9}{8} S_{2111}=\sum_{c y c} a^{2} c\left(a-c-\frac{3}{4} b+\frac{3}{4} d\right)^{2}, \\
S_{3200}-S_{2210}-S_{2201}+S_{3002}=\sum_{c y c} b^{2}\left(a^{3}-a^{2} c-a c^{2}+c^{3}\right)=\sum_{c y c} b^{2}(a+c)(a-c)^{2}, \\
S_{3200}-2 S_{3101}+S_{3002}=\sum_{c y c} a^{3}(b-d)^{2} \quad \text { and } \quad S_{3101}-S_{2111}=\frac{1}{3} \sum_{c y c} b d\left(2 a^{3}+c^{3}-3 a^{2} c\right)
\end{gathered}
$$

are all nonnegative.