Subfield: Combinatorics


Question: Given six positive numbers $a, b, c, d, e, f$ such that $a<b<c<d<e<f$. Let $a+c+e=S$ and $b+d+f=T$. Prove that

$$
2 S T>\sqrt{3(S+T)(S(b d+b f+d f)+T(a c+a e+c e))} .
\tag{1}
$$

Solution:
We define also $\sigma=a c+c e+a e, \tau=b d+b f+d f$. The idea of the solution is to interpret (1) as a natural inequality on the roots of an appropriate polynomial.

Actually, consider the polynomial

$$
\begin{array}{r}
P(x)=(b+d+f)(x-a)(x-c)(x-e)+(a+c+e)(x-b)(x-d)(x-f) \\
=T\left(x^{3}-S x^{2}+\sigma x-a c e\right)+S\left(x^{3}-T x^{2}+\tau x-b d f\right) .
\end{array}
\tag{2}
$$

Surely, $P$ is cubic with leading coefficient $S+T>0$. Moreover, we have

$$
\begin{array}{ll}
P(a)=S(a-b)(a-d)(a-f)<0, & P(c)=S(c-b)(c-d)(c-f)>0 \\
P(e)=S(e-b)(e-d)(e-f)<0, & P(f)=T(f-a)(f-c)(f-e)>0
\end{array}
$$

Hence, each of the intervals $(a, c),(c, e),(e, f)$ contains at least one root of $P(x)$. Since there are at most three roots at all, we obtain that there is exactly one root in each interval (denote them by $\alpha \in(a, c), \beta \in(c, e), \gamma \in(e, f))$. Moreover, the polynomial $P$ can be factorized as

$$
P(x)=(T+S)(x-\alpha)(x-\beta)(x-\gamma) \text {. }
\tag{3}
$$

Equating the coefficients in the two representations (2) and (3) of $P(x)$ provides

$$
\alpha+\beta+\gamma=\frac{2 T S}{T+S}, \quad \alpha \beta+\alpha \gamma+\beta \gamma=\frac{S \tau+T \sigma}{T+S}
$$

Now, since the numbers $\alpha, \beta, \gamma$ are distinct, we have

$$
0<(\alpha-\beta)^{2}+(\alpha-\gamma)^{2}+(\beta-\gamma)^{2}=2(\alpha+\beta+\gamma)^{2}-6(\alpha \beta+\alpha \gamma+\beta \gamma)
$$

which implies

$$
\frac{4 S^{2} T^{2}}{(T+S)^{2}}=(\alpha+\beta+\gamma)^{2}>3(\alpha \beta+\alpha \gamma+\beta \gamma)=\frac{3(S \tau+T \sigma)}{T+S}
$$

or

$$
4 S^{2} T^{2}>3(T+S)(T \sigma+S \tau)
$$

which is exactly what we need.
Let

$$
U=\frac{1}{2}\left((e-a)^{2}+(c-a)^{2}+(e-c)^{2}\right)=S^{2}-3(a c+a e+c e)
$$

and

$$
V=\frac{1}{2}\left((f-b)^{2}+(f-d)^{2}+(d-b)^{2}\right)=T^{2}-3(b d+b f+d f)
$$

Then

$$
\begin{aligned}
& \text { (L.H.S. })^{2}-(\text { R.H.S. })^{2}=(2 S T)^{2}-(S+T)(S \cdot 3(b d+b f+d f)+T \cdot 3(a c+a e+c e))= \\
& \quad=4 S^{2} T^{2}-(S+T)\left(S\left(T^{2}-V\right)+T\left(S^{2}-U\right)\right)=(S+T)(S V+T U)-S T(T-S)^{2},
\end{aligned}
$$

and the statement is equivalent with

$$
(S+T)(S V+T U)>S T(T-S)^{2} .\tag{4}
$$

By the Cauchy-Schwarz inequality,

$$
(S+T)(T U+S V) \geq(\sqrt{S \cdot T U}+\sqrt{T \cdot S V})^{2}=S T(\sqrt{U}+\sqrt{V})^{2} .
\tag{5}
$$

Estimate the quantities $\sqrt{U}$ and $\sqrt{V}$ by the QM-AM inequality with the positive terms $(e-c)^{2}$ and $(d-b)^{2}$ being omitted:

$$
\begin{aligned}
\sqrt{U}+\sqrt{V} & >\sqrt{\frac{(e-a)^{2}+(c-a)^{2}}{2}}+\sqrt{\frac{(f-b)^{2}+(f-d)^{2}}{2}} \\
& >\frac{(e-a)+(c-a)}{2}+\frac{(f-b)+(f-d)}{2}=\left(f-\frac{d}{2}-\frac{b}{2}\right)+\left(\frac{e}{2}+\frac{c}{2}-a\right) \\
& =(T-S)+\frac{3}{2}(e-d)+\frac{3}{2}(c-b)>T-S .
\end{aligned}\tag{6}
$$

The estimates (5) and (6) prove (4) and hence the statement.
We define $\sigma=a c+c e+a e, \tau=b d+b f+d f$.
Moreover, let $s=c+e$. Note that

$$
(c-b)(c-d)+(e-f)(e-d)+(e-f)(c-b)<0
$$

since each summand is negative. This rewrites as

$$
\begin{gathered}
(b d+b f+d f)-(a c+c e+a e)<(c+e)(b+d+f-a-c-e), \text { or } \\
\tau-\sigma<s(T-S) .
\end{gathered}
\tag{7}
$$

Then we have

$$
\begin{aligned}
S \tau+T \sigma & =S(\tau-\sigma)+(S+T) \sigma<S s(T-S)+(S+T)(c e+a s) \\
& \leq S s(T-S)+(S+T)\left(\frac{s^{2}}{4}+(S-s) s\right)=s\left(2 S T-\frac{3}{4}(S+T) s\right)
\end{aligned}
$$

Using this inequality together with the AM-GM inequality we get

$$
\begin{aligned}
\sqrt{\frac{3}{4}(S+T)(S \tau+T \sigma)} & <\sqrt{\frac{3}{4}(S+T) s\left(2 S T-\frac{3}{4}(S+T) s\right)} \\
& \leq \frac{\frac{3}{4}(S+T) s+2 S T-\frac{3}{4}(S+T) s}{2}=S T .
\end{aligned}
$$

Hence,

$$
2 S T>\sqrt{3(S+T)(S(b d+b f+d f)+T(a c+a e+c e))} .
$$
We introduce the expressions $\sigma$ and $\tau$ as in the previous solutions. The idea of the solution is to change the values of variables $a, \ldots, f$ keeping the left-hand side unchanged and increasing the right-hand side; it will lead to a simpler inequality which can be proved in a direct way.

Namely, we change the variables (i) keeping the (non-strict) inequalities $a \leq b \leq c \leq d \leq$ $e \leq f$; (ii) keeping the values of sums $S$ and $T$ unchanged; and finally (iii) increasing the values of $\sigma$ and $\tau$. Then the left-hand side of (1) remains unchanged, while the right-hand side increases. Hence, the inequality (1) (and even a non-strict version of (1)) for the changed values would imply the same (strict) inequality for the original values.

First, we find the sufficient conditions for (ii) and (iii) to be satisfied.

Lemma. Let $x, y, z>0$; denote $U(x, y, z)=x+y+z, v(x, y, z)=x y+x z+y z$. Suppose that $x^{\prime}+y^{\prime}=x+y$ but $|x-y| \geq\left|x^{\prime}-y^{\prime}\right|$; then we have $U\left(x^{\prime}, y^{\prime}, z\right)=U(x, y, z)$ and $v\left(x^{\prime}, y^{\prime}, z\right) \geq$ $v(x, y, z)$ with equality achieved only when $|x-y|=\left|x^{\prime}-y^{\prime}\right|$.

Proof. The first equality is obvious. For the second, we have

$$
\begin{aligned}
v\left(x^{\prime}, y^{\prime}, z\right)=z\left(x^{\prime}+y^{\prime}\right)+x^{\prime} y^{\prime} & =z\left(x^{\prime}+y^{\prime}\right)+\frac{\left(x^{\prime}+y^{\prime}\right)^{2}-\left(x^{\prime}-y^{\prime}\right)^{2}}{4} \\
& \geq z(x+y)+\frac{(x+y)^{2}-(x-y)^{2}}{4}=v(x, y, z),
\end{aligned}
$$

with the equality achieved only for $\left(x^{\prime}-y^{\prime}\right)^{2}=(x-y)^{2} \Longleftrightarrow\left|x^{\prime}-y^{\prime}\right|=|x-y|$, as desired.

Now, we apply Lemma several times making the following changes. For each change, we denote the new values by the same letters to avoid cumbersome notations.

1. Let $k=\frac{d-c}{2}$. Replace $(b, c, d, e)$ by $(b+k, c+k, d-k, e-k)$. After the change we have $a<b<c=d<e<f$, the values of $S, T$ remain unchanged, but $\sigma, \tau$ strictly increase by Lemma.
2. Let $\ell=\frac{e-d}{2}$. Replace $(c, d, e, f)$ by $(c+\ell, d+\ell, e-\ell, f-\ell)$. After the change we have $a<b<c=d=e<f$, the values of $S, T$ remain unchanged, but $\sigma, \tau$ strictly increase by the Lemma.
3. Finally, let $m=\frac{c-b}{3}$. Replace $(a, b, c, d, e, f)$ by $(a+2 m, b+2 m, c-m, d-m, e-m, f-m)$. After the change, we have $a<b=c=d=e<f$ and $S, T$ are unchanged. To check (iii), we observe that our change can be considered as a composition of two changes: $(a, b, c, d) \rightarrow$ $(a+m, b+m, c-m, d-m)$ and $(a, b, e, f) \rightarrow(a+m, b+m, e-m, f-m)$. It is easy to see that each of these two consecutive changes satisfy the conditions of the Lemma, hence the values of $\sigma$ and $\tau$ increase.

Finally, we come to the situation when $a<b=c=d=e<f$, and we need to prove the inequality

$$
\begin{aligned}
2(a+2 b)(2 b+f) & \geq \sqrt{3(a+4 b+f)\left((a+2 b)\left(b^{2}+2 b f\right)+(2 b+f)\left(2 a b+b^{2}\right)\right)} \\
& =\sqrt{3 b(a+4 b+f) \cdot((a+2 b)(b+2 f)+(2 b+f)(2 a+b)) .}
\end{aligned}
\tag{8}
$$

Now, observe that

$$
2 \cdot 2(a+2 b)(2 b+f)=3 b(a+4 b+f)+((a+2 b)(b+2 f)+(2 a+b)(2 b+f)) .
$$



Hence (4) rewrites as

$$
\begin{aligned}
3 b(a+4 b+f) & +((a+2 b)(b+2 f)+(2 a+b)(2 b+f)) \\
& \geq 2 \sqrt{3 b(a+4 b+f) \cdot((a+2 b)(b+2 f)+(2 b+f)(2 a+b))}
\end{aligned}
$$

which is simply the AM-GM inequality.