IMO 1995 Problem 2

Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 7m54s

Problem

Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $$ \frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} \geq \frac{3}{2}. $$

Exploration

The corrected target inequality is equivalent to proving a lower bound for a cyclic sum of fractions with denominators $a^3(b+c)$. The substitution $a=\frac1x$, $b=\frac1y$, $c=\frac1z$ with $xyz=1$ is structurally safe because it preserves positivity and converts multiplicative constraints into a symmetric one.

This transformation yields the expression

$\sum \frac{x^2}{y+z}, \quad xyz=1.$

A direct attempt to isolate each term using AM–GM on $y+z$ fails because replacing a sum in the denominator requires careful control of inequality direction; the earlier proof broke exactly here. A successful approach must avoid replacing denominators by weaker upper or lower bounds individually.

The expression suggests a global symmetrization strategy because each denominator appears twice in the full cyclic structure. This indicates that a weighted Cauchy–Schwarz or Titu-type approach may compress the sum into a single rational expression involving $x+y+z$ and $xy+yz+zx$.

Testing the symmetric point $x=y=z=1$ gives value $\frac{3}{2}$, so any correct argument must naturally produce $(1,1,1)$ as the equality case without artificial forcing.

No counterexample appears under symmetric perturbations such as $(2,1,\frac12)$, since direct substitution keeps the expression stable above $1.5$ in numerical checks, consistent with a global inequality driven by convexity in the reciprocal structure.

Problem Understanding

Let $a,b,c$ be positive real numbers satisfying $abc=1$. The goal is to prove

$\frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} \ge \frac{3}{2}.$

The expression is cyclic and homogeneous under scaling compatible with $abc=1$. Equality is expected at $a=b=c=1$, where each term equals $\frac12$ and the total equals $\frac32$.

The substitution $x=\frac1a$, $y=\frac1b$, $z=\frac1c$ converts the condition into $xyz=1$ and transforms the expression into

$\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}.$

The task reduces to proving a symmetric inequality in $(x,y,z)$ under a fixed geometric mean constraint.

Key Observations

Each denominator such as $y+z$ appears in a form suited for a single application of the Cauchy–Schwarz inequality in Titu’s lemma form:

$\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y} \ge \frac{(x+y+z)^2}{(y+z)+(z+x)+(x+y)}.$

The denominator simplifies because each variable appears exactly twice, giving

$(y+z)+(z+x)+(x+y)=2(x+y+z).$

This reduces the expression to a linear form in $x+y+z$, which can then be bounded from below using the constraint $xyz=1$ via AM–GM.

The structure shows that the earlier attempt failed because it tried to estimate each term separately instead of exploiting the global cancellation in the sum of denominators.

Solution

Set $x=\frac1a$, $y=\frac1b$, $z=\frac1c$. The condition $abc=1$ becomes $xyz=1$. The expression becomes

$\sum \frac{x^2}{y+z}.$

Applying Titu’s lemma (Cauchy–Schwarz in Engel form),

$\sum \frac{x^2}{y+z} \ge \frac{(x+y+z)^2}{(y+z)+(z+x)+(x+y)}.$

The denominator simplifies as

$(y+z)+(z+x)+(x+y)=2(x+y+z).$

Hence,

$\sum \frac{x^2}{y+z} \ge \frac{(x+y+z)^2}{2(x+y+z)}=\frac{x+y+z}{2}.$

It remains to bound $x+y+z$ under $xyz=1$. By AM–GM,

$x+y+z \ge 3\sqrt[3]{xyz}=3.$

Therefore,

$\sum \frac{x^2}{y+z} \ge \frac{x+y+z}{2} \ge \frac{3}{2}.$

Substituting back gives

$\frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} \ge \frac{3}{2}.$

This completes the proof. ∎

Verification of Key Steps

The substitution $x=\frac1a$, $y=\frac1b$, $z=\frac1c$ preserves positivity and transforms $abc=1$ into $xyz=1$ by direct multiplication, since $abc=\frac{1}{xyz}$.

The identity

$\frac{1}{a^3(b+c)}=\frac{x^2}{y+z}$

holds because $a=\frac1x$ gives $a^3=\frac{1}{x^3}$ and $b+c=\frac1y+\frac1z=\frac{y+z}{yz}$, producing cancellation with $xyz=1$.

Titu’s lemma applies because all denominators $y+z$, $z+x$, $x+y$ are positive. The aggregation step is exact since each variable occurs in exactly two denominators, yielding a precise factor of $2(x+y+z)$ with no approximation.

The AM–GM bound $x+y+z\ge 3$ is sharp under $xyz=1$, with equality only at $x=y=z=1$, which corresponds to $a=b=c=1$ and matches the equality case of the original expression.

No step reverses inequality direction or uses non-equivalent bounds on individual terms, so the argument remains globally valid.

Alternative Approaches

A second approach starts from the substitution $a=\frac1x$ and rewrites each term as

$\frac{x^2}{y+z}=\frac{x^2(x+y+z)}{(x+y+z)(y+z)}.$

Summing and applying rearrangement leads to symmetric expressions in $xy+yz+zx$, after which homogenization with $xyz=1$ reduces the problem to bounding a rational function in elementary symmetric polynomials.

Another route uses the substitution $a=\frac{y+z}{x}$ under cyclic consistency conditions, converting the inequality into a direct comparison between quadratic forms, followed by AM–GM on $x+y+z$.