Solve the following inequality:

$$
\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}>\frac{4}{3} \sqrt{\frac{x}{x+\sqrt{x}}}
$$
Solution. Under the square root, only non-negative numbers can stand, so the expressions only make sense if $x \geq 0, x-\sqrt{x} \geq 0$, that is, $x \geq 1$. In this case,

$$
x+\sqrt{x} \geq 0 \quad \text { and } \quad \frac{x}{x+\sqrt{x}} \geq 0
$$

are also satisfied. Therefore, $x \geq 1$.

Multiply both sides of the inequality by the positive

$$
\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}
$$

expression, and apply the identity $a^{2}-b^{2}=(a-b)(a+b)$ on the left side:

$$
\begin{aligned}
& (\sqrt{x+\sqrt{x}})^{2}-(\sqrt{x-\sqrt{x}})^{2}>\frac{4}{3} \sqrt{\frac{x}{x+\sqrt{x}}}(\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}) \\
& (x+\sqrt{x})-(x-\sqrt{x})>\frac{4}{3} \sqrt{x}+\frac{4}{3} \sqrt{x} \sqrt{\frac{x-\sqrt{x}}{x+\sqrt{x}}} \\
& 2 \sqrt{x}>\frac{4}{3} \sqrt{x}+\frac{4}{3} \sqrt{x} \sqrt{\frac{x-\sqrt{x}}{x+\sqrt{x}}}
\end{aligned}
$$

Rearranging and dividing by $\frac{4}{3} \sqrt{x}>0$, we get:

$$
\frac{1}{2}>\sqrt{\frac{x-\sqrt{x}}{x+\sqrt{x}}}
$$

Both sides are positive or zero, so squaring both sides results in an equivalent inequality:

$$
\frac{1}{4}>\frac{x-\sqrt{x}}{x+\sqrt{x}}
$$

Rearranging and dividing by $\sqrt{x}>0$:

$$
x+\sqrt{x}>4(x-\sqrt{x}), \quad 5 \sqrt{x}>3 x, \quad \frac{5}{3}>\sqrt{x}
$$

Squaring again: $\frac{25}{9}>x$.

Since we performed equivalent transformations due to the constraint $x \geq 1$, the inequality holds precisely when $1 \leq x<\frac{25}{9}$.