960*. Solve the inequality in natural numbers

$$
3^{x}+4^{x}<5^{x}
$$
$\triangle$ Calculations show that the inequality does not hold for $x=1$ and $x=2$, but it does hold for $x=3,4,5$.

We will prove that the inequality is valid for all natural $x \geqslant 3$. We will use the method of mathematical induction.

1) For $x=3$, the inequality is valid.
2) Assume that it is valid for some natural $x=k$, where $k \geqslant 3$:

$$
3^{k}+4^{k}<5^{k}
$$

We need to prove that it is also valid for $x=k+1$.

Rewrite the last inequality as:

$$
\left(\frac{3}{5}\right)^{k}+\left(\frac{4}{5}\right)^{k}<1
$$

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Which is greater, $\left(\frac{3}{5}\right)^{k}$ or $\left(\frac{3}{5}\right)^{k+1} ?\left(\frac{3}{5}\right)^{k}$, since multiplying the inequality $\frac{3}{5}<1$ by $\left(\frac{3}{5}\right)^{k}$, we get:

$$
\left(\frac{3}{5}\right)^{k+1}<\left(\frac{3}{5}\right)^{k}
$$

(If you are familiar with the exponential function, you can use its decreasing nature when the base is less than 1.)

Adding the inequalities

$$
\left(\frac{3}{5}\right)^{k+1}<\left(\frac{3}{5}\right)^{k}, \quad\left(\frac{4}{5}\right)^{k+1}<\left(\frac{4}{5}\right)^{k}
$$

we obtain:

$$
\left(\frac{3}{5}\right)^{k+1}+\left(\frac{4}{5}\right)^{k+1}<\left(\frac{3}{5}\right)^{k}+\left(\frac{4}{5}\right)^{k}<1
$$

Thus,

$$
\left(\frac{3}{5}\right)^{k+1}+\left(\frac{4}{5}\right)^{k+1}<1, \quad 3^{k+1}+4^{k+1}<5^{k+1}
$$

We have proven that if the original inequality is valid for $x=k \geqslant 3$, then it is also valid for $x=k+1$. Therefore, by the principle of mathematical induction, it is valid for any natural $x \geqslant 3$.

Answer: all natural $x$, not less than 3.