Kvant Math Problem 1262
Let the side lengths be $a,b,c$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 5m22s
Source on kvant.digital
Problem
Let $d_1$, $d_2$, $d_3$ be the pairwise differences of the side lengths of a triangle (in absolute value), and let $P$ be its perimeter. Prove the inequality $$d_1d_2+d_2d_3+d_3d_1\le\dfrac{P^2}{4}.$$
L. D. Kurlandchik
Exploration
Let the side lengths be $a,b,c$. The quantities $d_1,d_2,d_3$ are the three numbers
$$|a-b|,\quad |b-c|,\quad |c-a|.$$
The expression is symmetric in $d_1,d_2,d_3$, so the labeling is irrelevant. Since the sides of a triangle satisfy the triangle inequalities, it is natural to order them:
$$a\ge b\ge c>0.$$
Then
$$d_1=a-b,\qquad d_2=b-c,\qquad d_3=a-c.$$
Because
$$a-c=(a-b)+(b-c),$$
the three differences are not independent.
Write
$$x=a-b,\qquad y=b-c.$$
Then $d_3=x+y$, and
$$d_1d_2+d_2d_3+d_3d_1 =xy+y(x+y)+x(x+y) =x^2+3xy+y^2.$$
The perimeter is
$$P=a+b+c=(c+x)+(c+y)+c=3c+x+y.$$
The triangle condition $a<b+c$ becomes
$$c+x<c+(c+y),$$
hence
$$x<c+y.$$
This gives
$$c>x-y.$$
If $x\le y$, this is automatic; if $x>y$, it yields a useful lower bound on $c$.
To maximize the left-hand side for fixed $x,y$, one wants $c$ as small as possible because $P=3c+x+y$. The smallest admissible $c$ is approached by $c=x-y$ when $x\ge y$. Substituting this boundary value gives
$$P\to 4x-2y.$$
Then
$$\frac{P^2}{4}\to (2x-y)^2.$$
Comparing,
$$(2x-y)^2-(x^2+3xy+y^2) =3x^2-7xy.$$
This is not always nonnegative, so this line of attack is not immediately decisive.
A better idea is to express everything through
$$u=a-c,\qquad v=b-c.$$
Then $u\ge v\ge0$,
$$d_1=u-v,\qquad d_2=v,\qquad d_3=u,$$
and
$$d_1d_2+d_2d_3+d_3d_1 =(u-v)v+uv+u(u-v) =u^2+uv-v^2.$$
Also
$$P=(c+u)+(c+v)+c=3c+u+v.$$
The triangle inequality $a<b+c$ becomes
$$c+u<c+v+c,$$
hence
$$c>u-v.$$
Therefore
$$P>3(u-v)+u+v=4u-2v.$$
Consequently
$$\frac{P^2}{4}>(2u-v)^2.$$
Now compare:
$$(2u-v)^2-(u^2+uv-v^2) =3u^2-5uv+2v^2 =(u-v)(3u-2v)\ge0.$$
Since $u\ge v\ge0$, this works perfectly. This appears to be the key step.
Problem Understanding
We are given a triangle with side lengths $a,b,c$. Let $d_1,d_2,d_3$ be the absolute differences of the side lengths taken pairwise. The task is to prove that
$$d_1d_2+d_2d_3+d_3d_1\le \frac{P^2}{4},$$
where $P=a+b+c$ is the perimeter.
This is a Type B problem, a pure proof.
The core difficulty is to exploit the dependence among the three side differences imposed by the triangle inequalities. The crucial observation is that after ordering the sides and introducing suitable variables, the triangle inequality provides a lower bound for the perimeter, which can then be compared directly with the expression involving the differences.
Proof Architecture
Let $a\ge b\ge c$ and define $u=a-c$ and $v=b-c$; then $u\ge v\ge0$.
The first lemma is that
$$d_1=u-v,\qquad d_2=v,\qquad d_3=u,$$
hence
$$d_1d_2+d_2d_3+d_3d_1=u^2+uv-v^2.$$
This follows by direct substitution.
The second lemma is that the triangle inequality implies
$$c>u-v.$$
Indeed, $a<b+c$ becomes $c+u<c+v+c$.
The third lemma is that
$$P>4u-2v.$$
Substituting the previous inequality into $P=3c+u+v$ yields this estimate.
The final comparison is
$$(2u-v)^2-(u^2+uv-v^2) =(u-v)(3u-2v)\ge0.$$
Since $u\ge v\ge0$, the right-hand side is nonnegative.
The lemma most likely to fail under scrutiny is the last comparison; it must be factored correctly and checked using the ordering $u\ge v$.
Solution
Let the side lengths of the triangle be ordered as
$$a\ge b\ge c>0.$$
Define
$$u=a-c,\qquad v=b-c.$$
Then
$$u\ge v\ge0.$$
The pairwise differences of the side lengths are
$$a-b=u-v,\qquad b-c=v,\qquad a-c=u.$$
Hence
$$d_1d_2+d_2d_3+d_3d_1 =(u-v)v+uv+u(u-v).$$
Expanding and collecting terms gives
$$d_1d_2+d_2d_3+d_3d_1 =u^2+uv-v^2.$$
Next, use the triangle inequality $a<b+c$. Since $a=c+u$ and $b=c+v$,
$$c+u<c+v+c,$$
so
$$c>u-v.$$
The perimeter is
$$P=a+b+c=(c+u)+(c+v)+c=3c+u+v.$$
Using $c>u-v$,
$$P>3(u-v)+u+v=4u-2v.$$
Therefore
$$\frac{P^2}{4}>(2u-v)^2.$$
It remains to compare $(2u-v)^2$ with the left-hand side:
\begin{align*}
(2u-v)^2-(u^2+uv-v^2)
&=4u^2-4uv+v^2-u^2-uv+v^2\
&=3u^2-5uv+2v^2\
&=(u-v)(3u-2v).
\end{align*}
Since $u\ge v\ge0$, both factors on the right are nonnegative. Hence
$$(2u-v)^2\ge u^2+uv-v^2.$$
Combining this with $\frac{P^2}{4}>(2u-v)^2$, we obtain
$$\frac{P^2}{4}>u^2+uv-v^2 =d_1d_2+d_2d_3+d_3d_1.$$
Thus
$$d_1d_2+d_2d_3+d_3d_1\le\frac{P^2}{4}.$$
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the translation of the triangle inequality. Starting from
$$a<b+c,$$
substituting $a=c+u$ and $b=c+v$ gives
$$c+u<2c+v.$$
Subtracting $c+v$ from both sides yields
$$u-v<c,$$
which is exactly the required estimate. No other triangle inequality gives a stronger bound in these variables.
The second delicate step is the algebraic identity
$$(2u-v)^2-(u^2+uv-v^2) =(u-v)(3u-2v).$$
Expanding the factorization,
$$(u-v)(3u-2v)=3u^2-5uv+2v^2,$$
which matches the direct expansion of the left-hand side. A sign error in the term $-v^2$ would destroy the factorization.
The third delicate step is the sign check. From $u\ge v\ge0$,
$$u-v\ge0,$$
and
$$3u-2v\ge 3v-2v=v\ge0.$$
Hence the product is nonnegative. This uses both inequalities $u\ge v$ and $v\ge0$.
Alternative Approaches
One may instead set
$$x=a-b,\qquad y=b-c,$$
so that $a-c=x+y$. Then
$$d_1d_2+d_2d_3+d_3d_1=x^2+3xy+y^2.$$
The triangle inequality yields $c>x-y$, and since
$$P=3c+x+y,$$
one obtains
$$P>4x-2y.$$
The proof then reduces to verifying
$$(2x-y)^2-(x^2+3xy+y^2) =(x-y)(3x-2y)\ge0,$$
after choosing the ordering so that $x\ge y$.
The approach using $u=a-c$ and $v=b-c$ is preferable because the variables are automatically nonnegative and correspond directly to the two larger sides measured from the smallest side. The triangle inequality and the final factorization both emerge in a particularly transparent form.