How to Prove the Inequality \(\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}\geq\frac1a+\frac1b+\frac1c\)

Assume $a,b,c>0$, so every denominator is defined and multiplying an inequality by $abc$ preserves its direction. Under that standard assumption, the claimed strict inequality is not true for every choice: when $a=b=c$, the two sides are equal. The correct general result is a non-strict inequality, and the proof takes just two sum-of-squares comparisons.

The correct statement

For all positive real numbers $a,b,c$,

$$\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}\geq \frac1a+\frac1b+\frac1c.$$

Equality holds if and only if $a=b=c$. Therefore, the strict version is valid when $a,b,c$ are positive and not all equal.

Clear the denominators

Since $abc>0$, multiply both sides by it. Each term simplifies as follows:

$$\begin{aligned} abc\left(\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}\right) &=\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2},\\ abc\left(\frac1a+\frac1b+\frac1c\right) &=bc+ca+ab. \end{aligned}$$

So it remains to prove

$$\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}\geq ab+bc+ca.$$

Now define three positive quantities

$$x=\frac{ab}{c},\qquad y=\frac{bc}{a},\qquad z=\frac{ca}{b}.$$

The left side is $x^2+y^2+z^2$. The pairwise products on the right satisfy $xy=b^2$, $yz=c^2$, and $zx=a^2$. Thus the needed inequality follows from

$$x^2+y^2+z^2\geq xy+yz+zx=a^2+b^2+c^2\geq ab+bc+ca.$$

Both comparisons are standard: for any real $u,v,w$,

$$u^2+v^2+w^2-uv-vw-wu=\frac12\bigl((u-v)^2+(v-w)^2+(w-u)^2\bigr)\geq0.$$

Use this identity first with $u=x,v=y,w=z$, and then with $u=a,v=b,w=c$. Because the original inequality was multiplied by a positive number, this proves the desired result.

Worked example

Take $a=2$, $b=3$, and $c=4$. The original left side is

$$\frac{6}{64}+\frac{12}{8}+\frac{8}{27} =\frac{3}{32}+\frac32+\frac{8}{27} =\frac{1633}{864},$$

while the right side is

$$\frac12+\frac13+\frac14=\frac{13}{12}=\frac{936}{864}.$$

The left side is greater. The transformed comparison makes the reason visible: $x=\frac32$, $y=6$, and $z=\frac83$, so

$$x^2+y^2+z^2>xy+yz+zx=4+16+9=29>ab+bc+ca=26.$$

The first comparison is strict because $x,y,z$ are not all equal; the second is strict because $a,b,c$ are not all equal.

Equality and strictness

The identity for $x,y,z$ is an equality exactly when $x=y=z$. The identity for $a,b,c$ is an equality exactly when $a=b=c$. In the original inequality, equality occurs precisely when $a=b=c$: if all three equal a positive number $t$, each of the three terms on either side is $1/t$, and both sides sum to $3/t$.

For instance, $a=b=c=1$ gives $3=3$, not a strict inequality. If a problem requires a strict $>$ sign, it must exclude this equality case by assuming that $a,b,c$ are not all equal.

Common mistakes

• Keeping a strict sign without checking equality. Substituting $a=b=c$ immediately shows that the stated strict version cannot hold for all positive inputs.
• Multiplying without checking the sign. The proof uses $abc>0$. If variables are allowed to be negative, the denominators may still be defined, but multiplying by $abc$ can reverse the inequality; the positive-variable result does not automatically apply.
• Miscomputing the cleared terms. For example, $abc\cdot\frac{ab}{c^3}=\frac{a^2b^2}{c^2}$, not $a^2b^2c^2$.
• Using AM–GM in the wrong direction. The proof needs the identity $x^2+y^2+z^2\geq xy+yz+zx$, not a comparison that assumes the variables have a particular ordering.