## Task $9 / 73$

Determine the minimum of the function $y=x^{10}-1-\sqrt{2 x^{10}}-1$ without using tools from differential calculus.
We introduce the parameter $t=\sqrt{2 x^{10}-1}$. Then we have

$$
x^{10}=\frac{t^{2}+1}{2} \quad ; \quad y=\frac{t^{2}+1}{2}-1-t=\frac{(t-1)^{2}-2}{2}
$$

Obviously, due to $(t-1)^{2} \geq 0$, the minimum is achieved for $t-1=0$, i.e., for $t=1$. This value is indeed possible, yielding $x_{1 ; 2}= \pm 1 ; y_{\text {min }}=-1$.