\arccos\frac1{\sqrt{1+x}}\\ \arccos\frac1{\sqrt{1+x}}>\arccos(1-x)\\\frac1{\sqrt{1+x}}>1-x\\\frac1{x+1}>1-2x+x^2\\(x^2-2x+1)(x+1)<1\\x^3-x^2-x+1-1<0\\x(x^2-x-1)=0\\x_1=0\\x^2-x-1=0\\D=1+4\cdot1\cdot1=5\\x_2=\frac{1+\sqrt5}2,\;\;\;x_2=\frac{1-\sqrt5}2\\x\left(x-\frac{1-\sqrt5}2\right)\left(x-\frac{1+\sqrt5}2\right)<0\\x\in\left(-\infty;\;\frac{1-\sqrt5}2\right)\cup\left(0;\;\frac{1+\sqrt5}2\right)" alt="arctg\alpha=\arccos\frac1{\sqrt{1+\alpha^2}}\\arctg\sqrt x>\arccos\frac1{\sqrt{1+x}}\\ \arccos\frac1{\sqrt{1+x}}>\arccos(1-x)\\\frac1{\sqrt{1+x}}>1-x\\\frac1{x+1}>1-2x+x^2\\(x^2-2x+1)(x+1)<1\\x^3-x^2-x+1-1<0\\x(x^2-x-1)=0\\x_1=0\\x^2-x-1=0\\D=1+4\cdot1\cdot1=5\\x_2=\frac{1+\sqrt5}2,\;\;\;x_2=\frac{1-\sqrt5}2\\x\left(x-\frac{1-\sqrt5}2\right)\left(x-\frac{1+\sqrt5}2\right)<0\\x\in\left(-\infty;\;\frac{1-\sqrt5}2\right)\cup\left(0;\;\frac{1+\sqrt5}2\right)" align="absmiddle" class="latex-formula">