<s>[INST] How to solve irrational inequality? Irrational inequality wolfram alpha
I want to find $x$ such that
$\sqrt{x-3}+(9-x)^{1/4}>\sqrt{3}$.
Yeah, I know the answer but I don't know how to find this answer
 [/INST] The existence of roots implies that $3\le x\le 9$. Put $y=(9-x)^{1/2}$. Then $0\le y\le \sqrt{6}$ and
$$\sqrt{6-y^2}+\sqrt{y}>\sqrt{3}$$
Squaring both sides we obtain 
$$6-y^2+y+2\sqrt{6y-y^3}>3$$
$$2\sqrt{6y-y^3}>y^2-y-3$$
Then $y^2-y-3\le 0$ or $4(6y-y^3)>(y^2-y-3)^2$.


*

*$y^2-y-3\le 0$ implies $\frac{1-\sqrt{13}}2<0\le y\le\frac{1+\sqrt{13}}2<\sqrt{6}$. 

*$4(6y-y^3)>(y^2-y-3)^2$ implies $f(y)<0$, where $f(y)=y^4+2y^3-5y^2-18y+9$. 
The graph of the function $f(y)$ at the segment $\left[\frac{1+\sqrt{13}}2;\sqrt{6}\right]$ 

suggests  that an equation $f(y)=0$ has a unique root $y_0\simeq 2.44366$ at this segment. 
So the answer is $0\le y<y_0$, which yields an answer of the initial inequality $$3.0285\simeq 9-y_0^2 <x\le 9.$$
The equation $f(x)=0$ has fourth degree, so, theoretically, it can be solved in radicals (see, for instance, “Курс высшей алгебры” Куроша), but practically it is too complicated (for instance, in my practice which holds for more than quarter of a century, I never used Ferrari algorithm by hand). A form of Wolfram Alpha’s answer also suggests that there is no simple solution. 
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