Problem 1. Solve in the set of real numbers:
a) $4^{x} \cdot 9^{\frac{1}{x}}+9^{x} \cdot 4^{\frac{1}{x}}+6^{x+\frac{1}{x}}=108$
b) $13 \cdot\left(3^{\lg x^{2}}+x^{\lg 4}\right) \leq\left(3 \cdot x^{\lg 3}+2^{1+\lg x}\right)^{2}$
Solution:

a) Condition of existence: $x \neq 0$

For $x > 0$, the left-hand side can be written as $6^{x+\frac{1}{x}} \cdot\left[\left(\frac{2}{3}\right)^{x-\frac{1}{x}}+\left(\frac{3}{2}\right)^{x-\frac{1}{x}}+1\right]$.

But $x+\frac{1}{x} \geq 2 \Rightarrow 6^{x+\frac{1}{x}} \geq 36$.

Since $\left(\frac{2}{3}\right)^{x-\frac{1}{x}}+\left(\frac{3}{2}\right)^{x-\frac{1}{x}} \geq 2$, we deduce that the left-hand side is greater than or equal to 108. (2p)

Equality is obtained for $x=1$.

$1 p$
b) $\left(3 \cdot x^{\lg 3}+2^{\operatorname{l1} \lg x}\right)^{2}=\left(3 \cdot 3^{\lg x}+2 \cdot 2^{\lg x}\right)^{2} \leq 13 \cdot\left(3^{\lg x}+2^{2 \lg x}\right)$ according to C.B.S $1 p$

But $3^{2 \lg x}+2^{2 \lg x}=3^{\lg x^{2}}+x^{\lg 4}$.

Thus, the condition $\frac{3}{3^{\lg x}}=\frac{2}{2^{\lg x}} \Leftrightarrow\left(\frac{3}{2}\right)^{\lg x}=\frac{3}{2}$.

Therefore, $x=10$