Example 15. Solve the inequality

$$
2^{2 x^{2}-6 x+3}+6^{x^{2}-3 x+1} \geqslant 3^{2 x^{2}-6 x+3}
$$
Solution. Inequality (11) is equivalent to the inequality

$$
8 \cdot 2^{2\left(x^{2}-3 x\right)}+6 \cdot 2^{x^{2}-3 x} 3^{x^{2}-3 x}-27 \cdot 3^{2}\left(x^{2}-3 x\right) \geqslant 0,
$$

which is a homogeneous inequality of the form

$$
8 \cdot f^{2}(x)+6 f(x) \cdot g(x)-27 g^{2}(x) \geqslant 0
$$

where $f(x)=2^{x^{2}-3 x}, g(x)=3^{x^{2}-3 x}$.

Since $3^{x^{2}-3 x}>0$ for any $x$, dividing inequality (11) by $3^{2\left(x^{2}-3 x\right)}$ yields the equivalent inequality

$$
8 \cdot(2 / 3)^{2}\left(x^{2}-3 x\right)+6 \cdot(2 / 3)^{x^{2}-3 x}-27 \geqslant 0
$$

## § 3. EXPONENTIAL INEQUALITIES

The quadratic trinomial $8 y^{2}+6 y-27$ has roots $3 / 2$ and $-9 / 4$. Therefore, inequality (11) is equivalent to the system of inequalities

$$
\left(\frac{2}{3}\right)^{x^{2}-3 x} \leqslant-9 / 4, \quad\left(\frac{2}{3}\right)^{x^{2}-3 x} \geqslant 3 / 2
$$

The first inequality in this system has no solutions.

From the second inequality, we find:

$(2 / 3)^{x^{2}-3 x} \geqslant 3 / 2 \Leftrightarrow(2 / 3)^{x^{2}-3 x} \geqslant(2 / 3)^{-1} \Leftrightarrow$

$\Leftrightarrow x^{2}-3 x \leqslant-1 \Leftrightarrow x^{2}-3 x+1 \leqslant 0 \Leftrightarrow \frac{3-\sqrt{5}}{2} \leqslant x \leqslant \frac{3+\sqrt{5}}{2}$.

Thus, the set of all solutions to inequality (11) is the interval $\left[\frac{3-\sqrt{5}}{2} ; \frac{3+\sqrt{5}}{2}\right]$.