<s>[INST] Solve irrational inequality $\sqrt{x^2-8x+15} + \sqrt{x^2+2x-15} > \sqrt{4x^2-8x+18}$ Here's what I tried
$$\sqrt{x^2-8x+15} + \sqrt{x^2+2x-15} > \sqrt{4x^2-8x+18}$$
$$\sqrt{(x-3)(x-5)} + \sqrt{(x-3)(x+5)} > \sqrt{4x^2-8x+18}$$
Now,
$$(x-3)(x-5) > 0 \Leftrightarrow x \in (- \infty, 3) \cup (5, \infty)$$
$$(x-3)(x+5) > 0 \Leftrightarrow x \in (-\infty, -5) \cup (3, \infty)$$
$$4x^2-8x+18 > 0, \forall x \in R$$
So, for $x \in (- \infty, -5) \cup (5, \infty)$, we can take squares of both sides and we win this:
$$|x-3| \sqrt{(x-5)(x+5)} > x^2 - x + 9$$
Are my steps correct? And what should I do now? Should I solve for $x < -5,$ and for $x>5$?
Thanks.
 [/INST] You are correct so far. for $|x|\geq 5$ both sides of your last equation are positive, so you can square the equation again and solve the remaining equation (it is cubic at most). 
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