I would like to find without the use of derivatives of the minimum point of the function $$f (x) = 2 ^{- x-\sqrt {x ^ 2 + 1}} + 2 ^ {2x-2\sqrt {1-x ^ 2}}$$

(In fact, the point minimum is known: $x = 0$ and the minimum value of the function is equal to $3/4$).

Let $x=\sqrt{\cos 2y}, 0 \le 2y \le \frac{\pi}2$ $$f (x) = 2 ^{-\sqrt{\cos 2y} -\sqrt {1+\cos 2y}} + 2 ^ {2\sqrt{\cos 2y}-2\sqrt {1-{\cos 2y}}}$$

What's next?


Let $a,b > 0$. Then for any positive $t$ inequality $$at^2+\frac{b}t=at^2+\frac{b}{2t}+\frac{b}{2t}\ge3\sqrt[3]{\frac{ab^2}4}.$$

In our case we set $t=2^x$, $a=2^{-2\sqrt{1-x^2}}$, $b=2^{-\sqrt{1+x^2}}$. Then $ab^2=2^{-2(\sqrt{1-x^2}+\sqrt{1+x^2})}\ge\frac1{16}$ due to inequality $\sqrt{1-x^2}+\sqrt{1+x^2}\le2$, which obviously is checked by squaring.

Here $f(x)=at^2+\frac{b}t\ge3\sqrt[3]{\frac{ab^2}4}\ge\frac34$ and equality holds only at the origin.

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