When Is $b^2c^2+c^2a^2+a^2b^2>abc(a+b+c)$ True?

Assume first that $a,b,c$ are positive, as is usual for this kind of inequality. The strict inequality in the title needs an exception: when $a=b=c$, the two sides are equal. The correct general statement is a non-strict inequality, with strict inequality whenever the three variables are not all equal.

A direct sum-of-squares identity proves the result and identifies exactly when equality occurs.

Subtract the right-hand side

Set $x=bc$, $y=ca$, and $z=ab$. The left-hand side of the proposed inequality is $x^2+y^2+z^2$. Its right-hand side can also be written using these three quantities:

$$xy+yz+zx =abc(c+a+b) =abc(a+b+c).$$

Now use the identity

$$x^2+y^2+z^2-xy-yz-zx =\frac12\bigl((x-y)^2+(y-z)^2+(z-x)^2\bigr).$$

Substituting $x=bc$, $y=ca$, and $z=ab$ gives the exact difference between the two sides:

$$\begin{aligned} &b^2c^2+c^2a^2+a^2b^2-abc(a+b+c)\\ &\qquad=\frac12\bigl((bc-ca)^2+(ca-ab)^2+(ab-bc)^2\bigr)\\ &\qquad\ge 0. \end{aligned}$$

Each term on the final line is a square, so the difference cannot be negative. Therefore

$$b^2c^2+c^2a^2+a^2b^2\ge abc(a+b+c).$$

When is the inequality strict?

For equality, all three squares in the identity must be zero. That means

$$bc=ca=ab.$$

If $a,b,c>0$, divide these equalities by the positive variables: $bc=ca$ gives $b=a$, and $ca=ab$ gives $c=b$. Thus equality holds exactly when $a=b=c$.

Consequently, for positive variables the strict version is valid precisely when the variables are not all equal:

$$b^2c^2+c^2a^2+a^2b^2>abc(a+b+c) \quad\text{if }a,b,c>0\text{ and not all equal.}$$

At $a=b=c=t$ with $t>0$, both sides are $3t^4$. In particular, $a=b=c=1$ is a counterexample to the claim that the inequality is always strict.

A worked example

Take $a=1$, $b=2$, and $c=3$. The left side is

$$b^2c^2+c^2a^2+a^2b^2=2^2\cdot3^2+3^2\cdot1^2+1^2\cdot2^2=36+9+4=49,$$

while the right side is

$$abc(a+b+c)=1\cdot2\cdot3\cdot(1+2+3)=36.$$

The difference is $13$. The sum-of-squares identity checks the same result:

$$\frac12\bigl((6-3)^2+(3-2)^2+(2-6)^2\bigr) =\frac12(9+1+16)=13.$$

Since the variables are not all equal, the inequality is strict, as expected.

If the variables are not assumed positive

The sum-of-squares identity itself does not require positivity: it holds for all real $a,b,c$. Thus the non-strict inequality remains true for arbitrary real variables. Equality occurs whenever $ab=bc=ca$; this includes $a=b=c$, and also the cases where at least two variables are zero. For positive variables, the equality condition simplifies to $a=b=c$.

Positivity matters when describing equality, not when establishing nonnegativity: the proof comes directly from squares and does not require dividing by a variable.

Common mistakes

• Claiming strict inequality without checking equality. Any statement using $>$ must exclude $a=b=c$; otherwise use $\ge$.
• Applying AM-GM without connecting the terms. AM-GM can be useful in inequality proofs, but here the direct identity is shorter and gives the equality condition immediately.
• Dividing by a variable without knowing it is nonzero. The sum-of-squares proof works even when variables may be zero, so no division is needed.