Is This Inequality Always True for Positive x, y, z?

The inequality

$$\frac{x}{x+yz}+\frac{y}{y+zx}+\frac{z}{z+xy}\geq\frac{3}{2}$$

cannot be proved for arbitrary positive $x,y,z$: it is false without an additional condition. The key issue is that the terms compare variables with products, so scaling all three variables changes the fractions. Below are counterexamples and a simple sufficient condition under which the claimed bound does hold.

A counterexample for positive variables

Take $x=y=z=2$. Each fraction is

$$\frac{2}{2+2\cdot 2}=\frac{1}{3},$$

so the left side is $1$, which is less than $3/2$. Thus positivity alone is not enough.

In fact, setting $x=y=z=t$ gives

$$\frac{x}{x+yz}+\frac{y}{y+zx}+\frac{z}{z+xy} =\frac{3}{1+t}.$$

This is at least $3/2$ only when $t\leq 1$. For example, it equals $3/2$ at $t=1$, but is less than $3/2$ for every $t>1$. This also shows why a scaling argument cannot establish the inequality for all positive values: replacing each variable by a common multiple changes the result.

The condition $xyz=1$ is not enough

A common possible restriction is $xyz=1$, but that condition alone does not make the inequality true. For example, take $x=100$ and $y=z=0.1$. Their product is $1$, while

$$\begin{aligned} \frac{x}{x+yz}+\frac{y}{y+zx}+\frac{z}{z+xy} &=\frac{100}{100+0.01}+2\frac{0.1}{0.1+10}\\ &=\frac{10000}{10001}+\frac{2}{101}\\ &\approx 1.0197<\frac{3}{2}. \end{aligned}$$

So a proof that uses only $xyz=1$ must be missing another hypothesis or contain an error.

A simple sufficient condition

One useful condition that does make the inequality immediate is

$$x\geq yz,\qquad y\geq zx,\qquad z\geq xy.$$

Because the variables and denominators are positive, $x\geq yz$ implies $x+yz\leq 2x$, and hence

$$\frac{x}{x+yz}\geq\frac{1}{2}.$$

Applying the same argument to the other two inequalities gives

$$\frac{y}{y+zx}\geq\frac{1}{2},\qquad \frac{z}{z+xy}\geq\frac{1}{2}.$$

Adding these three bounds proves

$$\frac{x}{x+yz}+\frac{y}{y+zx}+\frac{z}{z+xy}\geq\frac{3}{2}.$$

For example, $x=y=z=1/2$ satisfies the three assumptions: each variable is at least the product of the other two. Each fraction is then $2/3$, and their sum is $2$. Equality in this proof requires all three comparisons $x\geq yz$, $y\geq zx$, and $z\geq xy$ to be equalities. For positive variables, that forces $x=y=z=1$.

This is a sufficient condition, not a claim that it is necessary. The inequality may also hold for values that do not satisfy all three comparisons.

Common mistakes

• Assuming positivity is enough. The equal-variable example $x=y=z=2$ disproves that immediately.
• Treating the expression as homogeneous. The numerator of each fraction has degree 1, while the product term in its denominator has degree 2. Scaling all variables therefore does not leave the expression unchanged.
• Using $xyz=1$ without checking the conclusion. The asymmetric example above has product $1$ but gives a sum below $3/2$.
• Applying a symmetric inequality without matching its hypotheses. Before using AM–GM, Muirhead, or a sum-of-squares argument, verify that the claimed inequality is actually true under the stated assumptions.

If a version of the problem includes a domain restriction, use that restriction explicitly. Without one, the correct response is that the proposed inequality is false, not to search for a proof of it.