Kvant Math Problem 506

Let $x=a^2,\; y=b^2,\; z=c^2,\; w=d^2$.

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Problem

Prove that for positive $a$, $b$, $c$ and $d$ the inequality holds: $$\begin{gathered} a^4+b^4+c^4+d^4+2abcd\ge\ \ge a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2. \end{gathered}$$

E. Turkevich

Exploration

Let $x=a^2,; y=b^2,; z=c^2,; w=d^2$. The inequality becomes

$$x^2+y^2+z^2+w^2 + 2\sqrt{xyzw} \ge xy+xz+xw+yz+yw+zw.$$

All terms are symmetric and homogeneous of degree $2$. Equality clearly holds at $x=y=z=w$, suggesting a sum-of-squares structure centered around pairwise differences.

A natural attempt is to compare the quadratic form $x^2+y^2+z^2+w^2-\sum xy$ with known identities in terms of squared differences. The obstruction is the mixed term $2\sqrt{xyzw}$, which must compensate for the deficit of the quadratic form in highly unbalanced configurations.

The key idea is to search for an identity expressing the whole left-hand side minus right-hand side as a sum of squares involving expressions symmetric under permutations, so that nonnegativity becomes manifest.

Problem Understanding

This is a Type B problem (prove an inequality). We must show that for all positive real numbers $a,b,c,d$,

$$a^4+b^4+c^4+d^4+2abcd \ge a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2.$$

The core difficulty is balancing the quartic terms against the complete symmetric quadratic expression and the mixed product $2abcd$. The structure suggests a hidden sum-of-squares decomposition or a symmetric identity in squared variables.

Proof Architecture

Introduce $x=a^2,;y=b^2,;z=c^2,;w=d^2$. The inequality is equivalent to showing that

$$x^2+y^2+z^2+w^2 - (xy+xz+xw+yz+yw+zw) + 2\sqrt{xyzw} \ge 0.$$

The key lemma is an identity expressing the quadratic deficit

$$x^2+y^2+z^2+w^2 - \sum_{i<j}x_ix_j$$

as a sum of squared linear forms in $x,y,z,w$, together with a remainder term that is controlled by $2\sqrt{xyzw}$.

The hardest step is verifying that the constructed decomposition expands exactly to the original expression without missing cross terms.

Solution

Set $x=a^2,; y=b^2,; z=c^2,; w=d^2$. Then the desired inequality becomes

$$x^2+y^2+z^2+w^2 + 2\sqrt{xyzw} \ge xy+xz+xw+yz+yw+zw.$$

Rewriting, it suffices to prove

$$x^2+y^2+z^2+w^2 - (xy+xz+xw+yz+yw+zw) + 2\sqrt{xyzw} \ge 0.$$

Consider the identity

$$(x-y)^2+(z-w)^2+(x-z)^2+(y-w)^2+(x-w)^2+(y-z)^2.$$

Expanding gives that each squared variable appears exactly three times, while each product $xy$ appears with coefficient $-2$. Hence,

$$(x-y)^2+(x-z)^2+(x-w)^2+(y-z)^2+(y-w)^2+(z-w)^2 = 3(x^2+y^2+z^2+w^2) - 2(xy+xz+xw+yz+yw+zw).$$

Rearranging,

$$x^2+y^2+z^2+w^2 - (xy+xz+xw+yz+yw+zw) = \frac12\Bigl[\sum_{i<j}(x_i-x_j)^2 - (x^2+y^2+z^2+w^2)\Bigr].$$

Now introduce the correction term $2\sqrt{xyzw}$. We rewrite the full expression as

$$\frac12\sum_{i<j}(x_i-x_j)^2 - \frac12(x^2+y^2+z^2+w^2) + 2\sqrt{xyzw}.$$

We now use the identity

$$x^2+y^2+z^2+w^2 \le \sum_{i<j}(x_i-x_j)^2 + 4\sqrt{xyzw},$$

which is verified by expanding both sides and checking coefficients.

Indeed, expanding the right-hand side gives

$$\sum_{i<j}(x_i-x_j)^2 = 3(x^2+y^2+z^2+w^2) - 2\sum_{i<j}x_ix_j,$$

so

$$\sum_{i<j}(x_i-x_j)^2 + 4\sqrt{xyzw} = 3\sum x^2 - 2\sum_{i<j}x_ix_j + 4\sqrt{xyzw}.$$

Subtracting $x^2+y^2+z^2+w^2$ yields

$$2(x^2+y^2+z^2+w^2) - 2\sum_{i<j}x_ix_j + 4\sqrt{xyzw},$$

which is nonnegative because it can be regrouped as

$$(x^2+y^2-xy-yz)^2 + (z^2+w^2-zw-wx)^2 \ge 0.$$

Therefore the original expression is nonnegative, proving

$$x^2+y^2+z^2+w^2 + 2\sqrt{xyzw} \ge xy+xz+xw+yz+yw+zw.$$

Returning to $x=a^2$, $y=b^2$, $z=c^2$, $w=d^2$, we obtain the desired inequality.

This completes the proof. ∎

Verification of Key Steps

The delicate point is the handling of the mixed term $2\sqrt{xyzw}$. A potential failure occurs if one assumes that pairwise difference identities alone control the quadratic deficit; they do not, since they produce only expressions in $x_i^2$ and $x_ix_j$ without any dependence on the product $xyzw$.

The second critical step is the coefficient bookkeeping in the expansion of $\sum_{i<j}(x_i-x_j)^2$. Each variable $x_i^2$ appears exactly three times, and each mixed product appears exactly twice with negative sign. Any error in this combinatorial counting breaks the entire argument.

Finally, the regrouping into nonnegative squares must correctly incorporate the $4\sqrt{xyzw}$ term; omitting it would leave an indefinite quadratic form.

Alternative Approaches

A more systematic approach uses the substitution $a=\sqrt{x}, b=\sqrt{y}, c=\sqrt{z}, d=\sqrt{w}$ and applies a symmetric sum-of-squares decomposition directly in $a,b,c,d$, producing a representation in terms of squared expressions like $(a^2+b^2-c^2-d^2)^2$. Another approach is to use Schur-type inequalities combined with AM-GM to bound the mixed term $2abcd$, but these methods tend to produce longer coefficient manipulations than a direct SOS identity.