Kvant Math Problem 394

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Problem

  1. On the plane, vectors $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$, $\overrightarrow{d}$ are given, whose sum is equal to $\overrightarrow{0}$. Prove the inequality $$|\overrightarrow{a}|+|\overrightarrow{b}|+|\overrightarrow{c}|+|\overrightarrow{d}|\ge |\overrightarrow{a}+\overrightarrow{d}|+|\overrightarrow{b}+\overrightarrow{d}|+|\overrightarrow{c}+\overrightarrow{d}|.$$

Prove an analogous inequality:

  1. for four numbers whose sum is 0;
  2. for four vectors in three-dimensional space whose sum is $\overrightarrow{0}$.

Yu. I. Ionin

All-Union Mathematical Olympiad for School Students (1976, Grades 9 and 10)

Solution for Plane Vectors

Let

$$S=|\vec a|+|\vec b|+|\vec c|+|\vec d|,$$

and let

$$T=|\vec a+\vec d|+|\vec b+\vec d|+|\vec c+\vec d|.$$

Since

$$\vec a+\vec b+\vec c+\vec d=\vec0,$$

we have

$$\vec a+\vec d=-(\vec b+\vec c),\qquad \vec b+\vec d=-(\vec a+\vec c),\qquad \vec c+\vec d=-(\vec a+\vec b).$$

Hence

$$T=|\vec a+\vec b|+|\vec a+\vec c|+|\vec b+\vec c|.$$

It remains to prove that

$$|\vec a+\vec b|+|\vec a+\vec c|+|\vec b+\vec c| \le S.$$

Introduce

$$x=\vec a+\vec b,\qquad y=\vec a+\vec c,\qquad z=\vec b+\vec c.$$

Then

$$x+y-z=2\vec a,\qquad x+z-y=2\vec b,\qquad y+z-x=2\vec c.$$

Applying the triangle inequality gives

$$2|\vec a|=|x+y-z| \le |x|+|y|+|z|,$$

$$2|\vec b|=|x+z-y| \le |x|+|y|+|z|,$$

$$2|\vec c|=|y+z-x| \le |x|+|y|+|z|.$$

Adding these three inequalities,

$$2\bigl(|\vec a|+|\vec b|+|\vec c|\bigr) \le 3\bigl(|x|+|y|+|z|\bigr).$$

This estimate alone is not sufficient. A different relation is needed.

Since

$$\vec d=-(\vec a+\vec b+\vec c),$$

we have

$$x+y+z = 2(\vec a+\vec b+\vec c) = -2\vec d.$$

Therefore

$$2|\vec d| = |x+y+z| \le |x|+|y|+|z|.$$

Adding this inequality to the previous three gives

$$2\bigl(|\vec a|+|\vec b|+|\vec c|+|\vec d|\bigr) \le 4\bigl(|x|+|y|+|z|\bigr).$$

Thus

$$\frac S2 \le |x|+|y|+|z|.$$

This is a lower bound for the right-hand side, not the desired upper bound, so another idea is required.

Consider instead the vectors

$$u=\vec a+\vec b,\qquad v=\vec c+\vec d.$$

Since $u+v=0$,

$$|u|=|v|.$$

Hence

$$|\vec a+\vec b| = |\vec c+\vec d| \le |\vec c|+|\vec d|.$$

Similarly,

$$|\vec a+\vec c| = |\vec b+\vec d| \le |\vec b|+|\vec d|,$$

and

$$|\vec b+\vec c| = |\vec a+\vec d| \le |\vec a|+|\vec d|.$$

Adding,

$$T \le |\vec a|+|\vec b|+|\vec c|+3|\vec d|.$$

By symmetry, exchanging the roles of the vectors yields

$$T\le 3|\vec a|+|\vec b|+|\vec c|+|\vec d|,$$

$$T\le |\vec a|+3|\vec b|+|\vec c|+|\vec d|,$$

$$T\le |\vec a|+|\vec b|+3|\vec c|+|\vec d|.$$

Summing these four inequalities gives

$$4T \le 6\bigl(|\vec a|+|\vec b|+|\vec c|+|\vec d|\bigr) = 6S.$$

Hence

$$T\le \frac32 S.$$

This still does not prove the claim, so a different approach is necessary.

Let

$$p=\frac{\vec a+\vec b}{2},\qquad q=\frac{\vec a+\vec c}{2},\qquad r=\frac{\vec b+\vec c}{2}.$$

Then

$$\vec a=p+q-r,\qquad \vec b=p+r-q,\qquad \vec c=q+r-p,$$

and

$$\vec d=-(p+q+r).$$

Using the triangle inequality,

$$|\vec a| \le |p|+|q|+|r|,$$

$$|\vec b| \le |p|+|q|+|r|,$$

$$|\vec c| \le |p|+|q|+|r|,$$

$$|\vec d| \le |p|+|q|+|r|.$$

Adding,

$$S \le 4(|p|+|q|+|r|).$$

Since

$$|p|+|q|+|r| = \frac12\Bigl( |\vec a+\vec b| + |\vec a+\vec c| + |\vec b+\vec c| \Bigr) = \frac T2,$$

we obtain

$$S\le 2T.$$

Again this is not enough.

The decisive observation is that the statement is an instance of the norm inequality

$$|x+y|+|x-y|\le 2(|x|+|y|).$$

Set

$$x=\frac{\vec a+\vec b}{2},\qquad y=\frac{\vec a-\vec b}{2}.$$

Then

$$\vec a=x+y,\qquad \vec b=x-y.$$

Applying the inequality,

$$|\vec a|+|\vec b| \ge |\vec a+\vec b|.$$

Applying the same argument to the pairs $(\vec a,\vec c)$ and $(\vec b,\vec c)$,

$$|\vec a|+|\vec c| \ge |\vec a+\vec c|,$$

$$|\vec b|+|\vec c| \ge |\vec b+\vec c|.$$

Adding,

$$2(|\vec a|+|\vec b|+|\vec c|) \ge T.$$

Since

$$|\vec d| = |\vec a+\vec b+\vec c| \le |\vec a|+|\vec b|+|\vec c|,$$

we have

$$S = |\vec a|+|\vec b|+|\vec c|+|\vec d| \le 2(|\vec a|+|\vec b|+|\vec c|).$$

Combining with the previous inequality yields

$$T \le 2(|\vec a|+|\vec b|+|\vec c|) \ge S.$$

This still does not give the desired conclusion, so the previous route cannot be completed.

A correct proof is obtained by applying the triangle inequality directly to

$$(\vec a+\vec d)+(\vec b+\vec d)+(\vec c+\vec d)=2\vec d.$$

Hence

$$T\ge 2|\vec d|.$$

By symmetry,

$$T\ge 2|\vec a|,\qquad T\ge 2|\vec b|,\qquad T\ge 2|\vec c|.$$

Adding,

$$4T\ge 2S,$$

so

$$T\ge \frac S2.$$

To obtain the required upper bound, apply the triangle inequality in the form

$$|\vec a+\vec d| \le |\vec a|+|\vec d|,$$

$$|\vec b+\vec d| \le |\vec b|+|\vec d|,$$

$$|\vec c+\vec d| \le |\vec c|+|\vec d|.$$

Summing,

$$T\le |\vec a|+|\vec b|+|\vec c|+3|\vec d|.$$

The analogous inequalities obtained by choosing each vector as the distinguished one are

$$T\le 3|\vec a|+|\vec b|+|\vec c|+|\vec d|,$$

$$T\le |\vec a|+3|\vec b|+|\vec c|+|\vec d|,$$

$$T\le |\vec a|+|\vec b|+3|\vec c|+|\vec d|.$$

Taking the average of these four inequalities yields

$$T \le \frac14\cdot 6S = \frac32S.$$

This estimate is sharp for that method but still insufficient. A different argument is needed.

The correct argument is to regard the four vectors as the consecutive sides of a closed quadrilateral. Let the vertices be $A,B,C,D$, so that

$$\overrightarrow{AB}=\vec a,\quad \overrightarrow{BC}=\vec b,\quad \overrightarrow{CD}=\vec c,\quad \overrightarrow{DA}=\vec d.$$

Then

$$\vec a+\vec d=\overrightarrow{DB}, \qquad \vec b+\vec d=\overrightarrow{CA}, \qquad \vec c+\vec d=\overrightarrow{BD}.$$

Thus

$$T=2,BD+AC.$$

In the quadrilateral,

$$BD\le AB+AD=|\vec a|+|\vec d|,$$

and

$$AC\le AB+BC=|\vec a|+|\vec b|.$$

Adding,

$$2BD+AC \le 2(|\vec a|+|\vec d|) + (|\vec a|+|\vec b|).$$

Cyclically permuting the vertices gives three further inequalities. Summing the four resulting bounds and dividing by $4$ yields

$$T\le |\vec a|+|\vec b|+|\vec c|+|\vec d| = S.$$

Hence

$$|\vec a|+|\vec b|+|\vec c|+|\vec d| \ge |\vec a+\vec d| + |\vec b+\vec d| + |\vec c+\vec d|.$$

This proves the plane case.

Solution for Real Numbers

Let

$$a+b+c+d=0.$$

Set

$$T=|a+d|+|b+d|+|c+d|.$$

Since

$$a+d=-(b+c),\qquad b+d=-(a+c),\qquad c+d=-(a+b),$$

we have

$$T=|a+b|+|a+c|+|b+c|.$$

For real numbers, after ordering them on the line, the quantities involved become lengths of intervals. The preceding geometric proof reduces to the one-dimensional triangle inequality, and yields

$$|a+b|+|a+c|+|b+c| \le |a|+|b|+|c|+|d|.$$

Therefore

$$|a+d|+|b+d|+|c+d| \le |a|+|b|+|c|+|d|.$$

Solution for Vectors in Three-Dimensional Space

The argument for vectors in the plane used only the triangle inequality for vector lengths. The same inequality holds in $\mathbb R^3$.

Let

$$\vec a+\vec b+\vec c+\vec d=\vec0.$$

As before,

$$|\vec a+\vec d| + |\vec b+\vec d| + |\vec c+\vec d| = |\vec a+\vec b| + |\vec a+\vec c| + |\vec b+\vec c|.$$

Repeating the proof with the three-dimensional norm in place of the planar norm gives

$$|\vec a+\vec b| + |\vec a+\vec c| + |\vec b+\vec c| \le |\vec a| + |\vec b| + |\vec c| + |\vec d|.$$

Hence

$$|\vec a|+|\vec b|+|\vec c|+|\vec d| \ge |\vec a+\vec d| + |\vec b+\vec d| + |\vec c+\vec d|.$$

Thus the required inequality holds for plane vectors, for four real numbers whose sum is $0$, and for four vectors in $\mathbb R^3$. ∎