Irrational Inequalities: Very Difficult Problems with Solutions

Solution:
Domain: [tex]x\ge 1\ \left(x^2-x+1\gt 0\right)[/tex].
Squaring: [tex]x^2-x+1\lt x+2\sqrt{x-1} \iff (x-1)^2\lt 2\sqrt{x-1}[/tex]
Substitute [tex]t=\sqrt{x-1}\ge 0[/tex]: [tex]t^4\lt 2t \iff t\left(t^3-2\right)\lt 0 \iff 0\lt t\lt \sqrt[3]{2}[/tex]
[tex]0\lt x-1\lt \sqrt[3]{4}[/tex]

Solution:
Domain: [tex]x\ge -\dfrac{1}{2}, x\ne 0[/tex].
Substitute [tex]s=\sqrt{1+2x}, 2x=s^2-1[/tex]: [tex]\dfrac{2x}{1-s}=\dfrac{(s-1)(s+1)}{1-s}=-(1+s)[/tex]
[tex]\dfrac{4x^2}{(1-s)^2}=(1+s)^2=2+2x+2s[/tex]
[tex]2+2x+2s\lt 2x+9 \iff s\lt \dfrac{7}{2} \iff 1+2x\lt \dfrac{49}{4} \iff x\lt \dfrac{45}{8}[/tex]
Answer: [tex]x\in\left[-\dfrac{1}{2}, 0\right)\cup\left(0, \dfrac{45}{8}\right)[/tex]

Solution:
Domain: [tex]x\ge -a[/tex].
[tex]x\lt -1:\ x+1\lt 0\ \checkmark[/tex] — [tex]x\in[-a, -1)\ne\varnothing \iff a\gt 1[/tex]
[tex]x\ge -1:\ 4(x+a)\gt (x+1)^2 \iff (x-1)^2\lt 4a \Rightarrow a\gt 0, 1-2\sqrt{a}\lt x\lt 1+2\sqrt{a}[/tex]
[tex]0\lt a\le 1:\ 1-2\sqrt{a}\ge -1 \Rightarrow x\in\left(1-2\sqrt{a}, 1+2\sqrt{a}\right)[/tex]
[tex]a\gt 1:\ x\in\left[-1, 1+2\sqrt{a}\right)\cup[-a, -1)=\left[-a, 1+2\sqrt{a}\right)[/tex]
Answer: [tex]a\le 0:\ x\in\varnothing\quad 0\lt a\le 1:\ x\in\left(1-2\sqrt{a}, 1+2\sqrt{a}\right)\quad a\gt 1:\ x\in\left[-a, 1+2\sqrt{a}\right)[/tex]

Solution:
[tex]x^2-8x+15=(x-3)(x-5), x^2+2x-15=(x-3)(x+5), 4x^2-18x+18=2(x-3)(2x-3)[/tex]
Domain: [tex]x\le -5, x=3[/tex] or [tex]x\ge 5[/tex].
[tex]x=3:\ 0\gt 0[/tex] — is false.
[tex]x\ge 5[/tex]. Dividing by [tex]\sqrt{x-3}[/tex]: [tex]\sqrt{x-5}+\sqrt{x+5}\gt \sqrt{4x-6} \iff \sqrt{x^2-25}\gt x-3 \iff x^2-25\gt x^2-6x+9 \iff x\gt \dfrac{17}{3}[/tex]
[tex]x\le -5[/tex]. Dividing by [tex]\sqrt{3-x}[/tex]: [tex]\sqrt{5-x}+\sqrt{-5-x}\gt \sqrt{6-4x} \iff \sqrt{x^2-25}\gt 3-x \Rightarrow x\gt \dfrac{17}{3}[/tex] — impossible.
Answer: [tex]x\in\left(\dfrac{17}{3}, +\infty\right)[/tex]

Solution:
Domain: [tex]x\ge \sqrt{2}[/tex]. By the AM–GM inequality:
[tex]\sqrt{2\left(x^2-2\right)}\le \dfrac{2+\left(x^2-2\right)}{2}=\dfrac{x^2}{2}, x\sqrt{1\times (x-1)}\le x\times \dfrac{1+(x-1)}{2}=\dfrac{x^2}{2}[/tex]
So the left side is at most [tex]x^2[/tex]; equality requires [tex]x^2-2=2[/tex] and [tex]x-1=1[/tex], i.e. [tex]x=2[/tex].
Answer: [tex]x=2[/tex]

Solution:
Domain: [tex]x^2\ge 5[/tex].
[tex]x^4-27x^2+200-6\sqrt{x^2-5}=\left(x^2-14\right)^2+\left(x^2-5\right)-6\sqrt{x^2-5}+9=\left(x^2-14\right)^2+\left(\sqrt{x^2-5}-3\right)^2\le 0[/tex]
[tex]x^2=14, \sqrt{14-5}=3\ \checkmark \Rightarrow x=\pm\sqrt{14}[/tex]

Solution:
Domain: [tex]-a\le x\le a, a\ge 0[/tex]. [tex]a=0:\ x=0, 0\gt 0[/tex] — is false.
[tex]a\gt 0[/tex]. Squaring: [tex]2a+2\sqrt{a^2-x^2}\gt a^2 \iff 2\sqrt{a^2-x^2}\gt a(a-2)[/tex]
[tex]0\lt a\lt 2:\ a(a-2)\lt 0 \Rightarrow x\in[-a, a][/tex]
[tex]a\ge 2:\ 4\left(a^2-x^2\right)\gt a^2(a-2)^2 \iff x^2\lt \dfrac{a^3(4-a)}{4}[/tex] — [tex]a\lt 4, |x|\lt \dfrac{a}{2}\sqrt{a(4-a)}\le a[/tex]
[tex]a\ge 4[/tex] — no solution.
Answer: [tex]a\notin(0, 4):\ x\in\varnothing\quad 0\lt a\lt 2:\ x\in[-a, a]\quad 2\le a\lt 4:\ |x|\lt \dfrac{a}{2}\sqrt{a(4-a)}[/tex]