Kvant Math Problem 1526

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Problem

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

International Mathematical Olympiad for School Students (XXXVI)

Exploration

The condition $abc=1$ suggests eliminating one variable from denominators by rewriting powers of $a,b,c$ in terms of the other two variables. The presence of cubic powers in the numerator structure $a^{-3}(b+c)^{-1}$ indicates that converting $1/a^3$ into a product of powers of $b$ and $c$ may produce a symmetric expression.

Writing $a=\frac{1}{bc}$ gives $a^{-3}=b^3c^3$, and each term becomes a rational function symmetric in cyclic shifts:

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

This reduces the problem to proving

$\sum \frac{b^3c^3}{b+c}\ge \frac{3}{2}.$

The structure suggests applying a weighted Cauchy inequality of the form $\sum \frac{x^2}{y}$, with $x$ chosen as a power of $bc$, followed by a reduction to a symmetric inequality in $a,b,c$. The main difficulty is controlling the resulting power sum $\sum (bc)^{3/2}$ under the constraint $abc=1$.

Problem Understanding

This is a Type C problem, requiring proof of a lower bound for a symmetric rational expression under the constraint $abc=1$.

After rewriting $a^{-3}$ as $b^3c^3$, the expression becomes symmetric in cyclic products $bc$, so the natural structure reduces the problem to bounding a sum depending only on pairwise products.

The expected extremum occurs at $a=b=c=1$, since the expression becomes symmetric and homogeneous after normalization, suggesting equality should hold at the balanced point.

Proof Architecture

The proof proceeds by rewriting the expression using $abc=1$ into a symmetric form $\sum \frac{b^3c^3}{b+c}$.

A first lemma applies Cauchy–Schwarz in the Engel form to compare $\sum \frac{b^3c^3}{b+c}$ with a quotient involving $\sum (bc)^{3/2}$.

A second lemma proves a lower bound on $\sum (bc)^{3/2}$ using the arithmetic–geometric mean inequality under the constraint $abc=1$.

A final combination of these estimates yields the desired bound.

The most delicate step is controlling $\sum (bc)^{3/2}$ under the multiplicative constraint.

Solution

Since $abc=1$, each variable can be written as $a=\frac{1}{bc}$, $b=\frac{1}{ca}$, and $c=\frac{1}{ab}$. It follows that

$\frac{1}{a^3(b+c)}=\frac{b^3c^3}{b+c},\quad \frac{1}{b^3(c+a)}=\frac{c^3a^3}{c+a},\quad \frac{1}{c^3(a+b)}=\frac{a^3b^3}{a+b}.$

The expression becomes

$\sum \frac{b^3c^3}{b+c}.$

For each term, apply the inequality

$\frac{x^2}{y}+\frac{z^2}{t}+\frac{u^2}{v}\ge \frac{(x+z+u)^2}{y+t+v}$

in the form of Cauchy–Schwarz:

$\sum \frac{b^3c^3}{b+c}=\sum \frac{(b^{3/2}c^{3/2})^2}{b+c}\ge \frac{\left(\sum b^{3/2}c^{3/2}\right)^2}{(b+c)+(c+a)+(a+b)}.$

The denominator simplifies to

$(b+c)+(c+a)+(a+b)=2(a+b+c),$

so

$\sum \frac{b^3c^3}{b+c}\ge \frac{\left(\sum b^{3/2}c^{3/2}\right)^2}{2(a+b+c)}.$

Now write $b^{3/2}c^{3/2}=(bc)^{3/2}$. Using $abc=1$, we have $bc=\frac{1}{a}$, hence

$(bc)^{3/2}=\frac{1}{a^{3/2}}.$

Therefore,

$\sum b^{3/2}c^{3/2}=\sum \frac{1}{a^{3/2}}.$

Apply the arithmetic–geometric mean inequality to the positive numbers $\frac{1}{a^{3/2}},\frac{1}{b^{3/2}},\frac{1}{c^{3/2}}$:

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

Since $a^{3/2}b^{3/2}c^{3/2}=(abc)^{3/2}=1$, this gives

$\sum \frac{1}{a^{3/2}}\ge 3.$

Substituting into the earlier estimate yields

$\sum \frac{b^3c^3}{b+c}\ge \frac{9}{2(a+b+c)}.$

It remains to bound $a+b+c$. By AM–GM,

$a+b+c\ge 3(abc)^{1/3}=3,$

so

$\frac{9}{2(a+b+c)}\ge \frac{9}{2\cdot 3}=\frac{3}{2}.$

Thus,

$\sum \frac{b^3c^3}{b+c}\ge \frac{3}{2},$

which is exactly the desired inequality.

This completes the proof. ∎

Verification of Key Steps

The reduction $a^{-3}=b^3c^3$ follows directly from $a=\frac{1}{bc}$ and cubing both sides, producing no hidden assumptions.

In the Cauchy–Schwarz step, the pairing uses nonnegative quantities $b^{3/2}c^{3/2}$, ensuring validity of the inequality in the form $\sum \frac{x_i^2}{y_i}\ge \frac{(\sum x_i)^2}{\sum y_i}$ with $y_i=b+c,c+a,a+b$.

The inequality $\sum \frac{1}{a^{3/2}}\ge 3$ is an equality case of AM–GM since the geometric mean equals $1$ under $abc=1$, forcing the arithmetic mean to be at least $1$.

Finally, the bound $a+b+c\ge 3$ is sharp at $a=b=c=1$, consistent with equality propagation through the chain of inequalities.

Alternative Approaches

One alternative method begins directly from the homogenized form $\sum \frac{b^3c^3}{b+c}$ and applies Hölder’s inequality in a single step to balance powers of $bc$ against linear terms $b+c$, avoiding the intermediate reduction to $\sum (bc)^{3/2}$. This approach compresses the argument but requires a more intricate choice of exponents to align homogeneity.

Another approach uses substitution $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$, transforming the inequality into a cyclic rational inequality in three independent variables, but the resulting expression is less symmetric and does not simplify as directly as the method based on pairwise products $bc$.