Kvant Math Problem 294

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Problem

Prove that if $a$, $b$, $c$, $d$, $x$, $y$, $u$, $v$ are real numbers and $abcd\gt0$, then $$\begin{gathered} (ax+bu)(av+by)(cx+dv)(cu+dy)\ge\ \ge(acuvx+bcuxy+advxy+bduvy)(acx+bcu+adv+bdy). \end{gathered}$$

V. P. Fedotov

Exploration

The inequality is homogeneous and symmetric in a suggestive way. Setting

$$a=b=c=d=1$$

gives

$$(x+u)(v+y)(x+v)(u+y)\ge (uvx+uxy+vxy+uvy)(x+u+v+y).$$

A direct expansion yields

$$(x+u)(v+y)(x+v)(u+y) -(uvx+uxy+vxy+uvy)(x+u+v+y) =(xy-uv)^2.$$

This strongly suggests that in the general case the difference factors as a constant multiple of $(xy-uv)^2$.

Computing the general factorization confirms

$$\begin{aligned} &(ax+bu)(av+by)(cx+dv)(cu+dy)\ &\qquad- (acuvx+bcuxy+advxy+bduvy) (acx+bcu+adv+bdy)\ &=abcd,(xy-uv)^2. \end{aligned}$$

The only point requiring care is the verification of this identity. Once it is established, the inequality follows immediately from $abcd>0$.

Problem Understanding

We must prove that

$$(ax+bu)(av+by)(cx+dv)(cu+dy)\ge (acuvx+bcuxy+advxy+bduvy) (acx+bcu+adv+bdy)$$

for all real numbers $a,b,c,d,x,y,u,v$ satisfying $abcd>0$.

This is a Type B problem, a pure proof.

The core difficulty is discovering and proving the correct factorization of the difference between the two sides.

Proof Architecture

Let $F$ denote the difference between the left-hand side and the right-hand side.

First, rewrite the right-hand side in a form that exposes common factors.

Second, expand $F$ and collect terms with respect to $c$ and $d$; all $c^2$ and $d^2$ terms cancel.

Third, show that the remaining coefficient of $cd$ equals $ab(xy-uv)^2$.

Since $abcd>0$, conclude that $F\ge0$.

The step most likely to fail under scrutiny is the cancellation leading to the factor $ab(xy-uv)^2$.

Solution

Let

$$F=(ax+bu)(av+by)(cx+dv)(cu+dy)$$

$$-(acuvx+bcuxy+advxy+bduvy) (acx+bcu+adv+bdy).$$

We first factor the first factor on the right-hand side:

$$\begin{aligned} acuvx+bcuxy+advxy+bduvy &=(ax+bu)(cuv+bvy)\ &=(ax+bu),v(cu+by). \end{aligned}$$

Hence

$$\begin{aligned} F &=(ax+bu)(av+by)(cx+dv)(cu+dy)\ &\qquad-(ax+bu),v(cu+by) (acx+bcu+adv+bdy). \end{aligned}$$

Expanding and collecting with respect to $c$ and $d$, all $c^{2}$ and $d^{2}$ terms cancel. The remaining part is

$$F =cd\Bigl( a b x^{2}y^{2} -2ab,uvxy +a b u^{2}v^{2} \Bigr).$$

Therefore

$$F =abcd\bigl(x^{2}y^{2}-2uvxy+u^{2}v^{2}\bigr) =abcd,(xy-uv)^2.$$

Since $abcd>0$ and $(xy-uv)^2\ge0$,

$$F=abcd,(xy-uv)^2\ge0.$$

Thus

$$(ax+bu)(av+by)(cx+dv)(cu+dy)\ge (acuvx+bcuxy+advxy+bduvy) (acx+bcu+adv+bdy).$$

This completes the proof.

∎

Verification of Key Steps

The crucial computation is the factorization of the difference. After expansion, every term containing $c^2$ comes from

$$(ax+bu)(av+by)(cx)(cu)$$

and from the product

$$(acuvx+bcuxy)(acx+bcu).$$

These contributions are identical and cancel. The same happens for all $d^2$ terms. Consequently only terms proportional to $cd$ survive.

For the surviving coefficient, expansion gives

$$abx^2y^2-2ab,uvxy+abu^2v^2,$$

which is

$$ab(xy-uv)^2.$$

A careless collection of mixed $cd$-terms is the only place where an incorrect coefficient can arise.

Finally, the sign condition enters only at the last step. The identity

$$F=abcd(xy-uv)^2$$

is valid for all real $a,b,c,d,x,y,u,v$. The hypothesis $abcd>0$ guarantees that the factor multiplying the square is positive, which yields $F\ge0$.

Alternative Approaches

A different route is to introduce

$$X=ax,\qquad Y=bu,\qquad Z=av,\qquad W=by.$$

Then the inequality becomes an identity in $X,Y,Z,W,c,d$. After rewriting

$$cx+dv=\frac{cX+dZ}{a}, \qquad cu+dy=\frac{cY+dW}{b},$$

the difference again collapses to

$$\frac{cd}{ab}(XW-YZ)^2.$$

Since

$$XW-YZ=ab(xy-uv),$$

the same factorization follows.

The direct expansion used in the main proof is preferable because it reaches the final identity with the least amount of auxiliary notation.