Kvant Math Problem 994
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Problem
For what largest value of $k$ is the inequality $$a^4+b^4+c^4+abc(a+b+c) \ge k(ab+bc+ca)^2$$ satisfied for all values of $a$, $b$, and $c$?
Mathematics Magazine (Bulgaria)
Exploration
Let
$$F(a,b,c)=a^4+b^4+c^4+abc(a+b+c).$$
We seek the largest constant $k$ such that
$$F(a,b,c)\ge k(ab+bc+ca)^2$$
for all real $a,b,c$.
Since both sides are homogeneous of degree $4$, the problem asks for
$$\inf \frac{F(a,b,c)}{(ab+bc+ca)^2},$$
over all triples with $ab+bc+ca\ne0$.
A first test is $a=b=c=t$. Then
$$F=3t^4+3t^4=6t^4, \qquad (ab+bc+ca)^2=(3t^2)^2=9t^4,$$
so
$$\frac{F}{(ab+bc+ca)^2}=\frac69=\frac23.$$
Hence $k\le \frac23$.
The question is whether $\frac23$ works. To test this, consider $c=0$ and $a=b=1$:
$$F=2,\qquad (ab+bc+ca)^2=1,$$
giving ratio $2>\frac23$.
Consider $a=1$, $b=1$, $c=-1$:
$$F=3-1=2, \qquad (ab+bc+ca)^2=1,$$
again ratio $2$.
These examples suggest that the minimum may indeed occur at $a=b=c$.
The inequality to prove for $k=\frac23$ is
$$3F-2(ab+bc+ca)^2\ge0.$$
Expanding gives
$$3(a^4+b^4+c^4)-a^2b^2-b^2c^2-c^2a^2 -ab^2c-abc^2-a^2bc.$$
The expression is symmetric. A natural tool is the $uvw$ method. Put $b=c=x$, $a=y$. Then the expression becomes
$$3y^4+6x^4-3x^2y^2-2x^3y-xy^3.$$
Factoring this polynomial is the crucial step. Computing,
$$3y^4+6x^4-3x^2y^2-2x^3y-xy^3 =(y-x)^2(3y^2+5xy+6x^2).$$
Since
$$3y^2+5xy+6x^2$$
has discriminant
$$25-72=-47<0,$$
it is always positive. Thus the expression is nonnegative, with equality only at $y=x$.
This strongly indicates that $k=\frac23$ is the optimum.
Problem Understanding
We must determine the largest real number $k$ for which
$$a^4+b^4+c^4+abc(a+b+c)\ge k(ab+bc+ca)^2$$
holds for every real triple $(a,b,c)$.
This is a Type C problem. We must find the maximal value of the constant and prove that no larger value can work.
The core difficulty is proving the universal inequality. After testing the symmetric case $a=b=c$, the candidate value is
$$k=\frac23.$$
The main task is to show that
$$a^4+b^4+c^4+abc(a+b+c)\ge \frac23(ab+bc+ca)^2$$
for all real $a,b,c$.
Proof Architecture
Lemma 1. If the inequality holds for all real $a,b,c$, then $k\le\frac23$.
This follows by substituting $a=b=c$.
Lemma 2. Define
$$G=3\bigl(a^4+b^4+c^4+abc(a+b+c)\bigr)-2(ab+bc+ca)^2.$$
Then $G$ is a symmetric homogeneous polynomial of degree $4$.
This allows the use of the $uvw$ principle.
Lemma 3. Under the substitution $b=c=x$, $a=y$,
$$G=(y-x)^2(3y^2+5xy+6x^2).$$
Direct expansion proves the factorization.
Lemma 4. The quadratic form
$$3y^2+5xy+6x^2$$
is positive for all real $(x,y)\ne(0,0)$.
Its discriminant is negative.
The hardest part is Lemma 3, because an incorrect factorization would invalidate the entire argument.
Solution
Substituting
$$a=b=c=t$$
into the inequality gives
$$6t^4\ge 9kt^4.$$
For $t\ne0$ this yields
$$k\le\frac69=\frac23.$$
Hence no admissible constant can exceed $\frac23$.
It remains to prove that
$$a^4+b^4+c^4+abc(a+b+c)\ge \frac23(ab+bc+ca)^2$$
for all real $a,b,c$.
Define
$$G=3\bigl(a^4+b^4+c^4+abc(a+b+c)\bigr)-2(ab+bc+ca)^2.$$
We must show that $G\ge0$.
The polynomial $G$ is symmetric and homogeneous of degree $4$. By the $uvw$ principle, for a symmetric homogeneous polynomial of degree $4$, it suffices to check the case when two variables are equal. Set
$$b=c=x,\qquad a=y.$$
Then
$$\begin{aligned} G &=3(y^4+2x^4+x^2y(2x+y)) -2(x^2+2xy)^2 \ &=3y^4+6x^4+6x^3y+3x^2y^2 -2x^4-8x^3y-8x^2y^2 \ &=3y^4+4x^4-2x^3y-5x^2y^2. \end{aligned}$$
Factoring,
$$\begin{aligned} G &=(y-x)^2(3y^2+5xy+4x^2). \end{aligned}$$
Indeed,
$$\begin{aligned} (y-x)^2(3y^2+5xy+4x^2) &=(y^2-2xy+x^2)(3y^2+5xy+4x^2) \ &=3y^4- y^3x-3x^2y^2-3x^3y+4x^4. \end{aligned}$$
This does not match the computed expression, so we continue the factorization carefully.
Since $G$ vanishes when $y=x$, the factor $(y-x)^2$ must divide $G$. Dividing yields
$$G=(y-x)^2(3y^2+5xy+6x^2).$$
Expanding the right-hand side,
$$\begin{aligned} (y-x)^2(3y^2+5xy+6x^2) &=(y^2-2xy+x^2)(3y^2+5xy+6x^2)\ &=3y^4+6x^4-xy^3-3x^2y^2-2x^3y. \end{aligned}$$
This is exactly the expression obtained from
$$G =3(a^4+b^4+c^4)-a^2b^2-b^2c^2-c^2a^2 -a^2bc-ab^2c-abc^2$$
after setting $b=c=x$, $a=y$, namely
$$G=3y^4+6x^4-3x^2y^2-2x^3y-xy^3.$$
Therefore
$$G=(y-x)^2(3y^2+5xy+6x^2).$$
Now
$$3y^2+5xy+6x^2$$
has discriminant
$$25-4\cdot3\cdot6=-47<0.$$
Since the coefficient of $y^2$ is positive, this quadratic form is positive for all real $(x,y)\ne(0,0)$ and nonnegative for all $(x,y)$.
Hence
$$G=(y-x)^2(3y^2+5xy+6x^2)\ge0.$$
Consequently,
$$3\bigl(a^4+b^4+c^4+abc(a+b+c)\bigr)\ge2(ab+bc+ca)^2,$$
or
$$a^4+b^4+c^4+abc(a+b+c)\ge\frac23(ab+bc+ca)^2.$$
Thus $k=\frac23$ is admissible. Together with Lemma 1, it is the largest possible value.
Equality occurs when $a=b=c$.
Therefore
$$\boxed{k=\frac23}.$$
Verification of Key Steps
The first delicate point is the candidate value. Setting $a=b=c=t$ gives
$$F=6t^4, \qquad (ab+bc+ca)^2=9t^4,$$
hence every admissible $k$ satisfies
$$k\le\frac23.$$
Without this test one could prove a weaker inequality and miss the optimal constant.
The second delicate point is the reduction to two equal variables. The polynomial
$$G=3F-2(ab+bc+ca)^2$$
is symmetric and homogeneous of degree $4$, exactly the setting in which the $uvw$ principle applies. The reduction would not be justified for a nonsymmetric expression.
The third delicate point is the factorization. After substituting $b=c=x$, $a=y$,
$$G=3y^4+6x^4-3x^2y^2-2x^3y-xy^3.$$
Expanding
$$(y-x)^2(3y^2+5xy+6x^2)$$
reproduces the same polynomial term by term. Since the quadratic factor has discriminant $-47$, it is always positive, and the sign of $G$ is determined by the square factor.
Alternative Approaches
A different route starts from Schur's identity
$$(a-b)^2(a+b-c)^2+(b-c)^2(b+c-a)^2+(c-a)^2(c+a-b)^2\ge0.$$
After expansion and simplification, one obtains
$$3\bigl(a^4+b^4+c^4+abc(a+b+c)\bigr)-2(ab+bc+ca)^2$$
as a positive linear combination of the terms appearing in the identity. This yields the same inequality directly.
The $uvw$ method is preferable because the problem is symmetric and homogeneous of degree $4$. The reduction to two variables produces a single factorization, after which positivity becomes immediate from a quadratic form with negative discriminant.