Maximize $P=\frac{\sqrt{a^2-1}}{a}+\frac{\sqrt{b^2-1}}{b}+\frac{\sqrt{c^2-1}}{c} $ Let $a,b,c\geq 1$ satisfy $ 32abc=18(a+b+c)+27$. Find the maximum value $$P=\dfrac{\sqrt{a^2-1}}{a}+\dfrac{\sqrt{b^2-1}}{b}+\dfrac{\sqrt{c^2-1}}{c} $$

$\sqrt {a^2-1}=\sqrt{(a-1)(a+1)}\leq \frac{a-1+a+1}{2}=a$
$\Rightarrow \frac{\sqrt {a^2-1}}{a}\leq \frac{\sqrt {a}}{a}=\frac{1}{\sqrt {a}}$
Need prove $\sum\frac{1}{\sqrt{a}}\leq \sqrt {5}$ ?
 A: If $a=b=c=\frac{3}{2}$ so $\sum\limits_{cyc}\frac{\sqrt{a^2-1}}{a}=\sqrt5.$
We'll prove that it's a maximal value.
Indeed, by C-S
$$\sum_{cyc}\frac{\sqrt{a^2-1}}{a}=\sum_{cyc}\sqrt{1-\frac{1}{a^2}}\leq\sqrt{(1+1+1)\sum_{cyc}\left(1-\frac{1}{a^2}\right)}=\sqrt{3\sum_{cyc}\left(1-\frac{1}{a^2}\right)}.$$
Thus, it remains to prove that:
$$3\sum_{cyc}\left(1-\frac{1}{a^2}\right)\leq5$$ or
$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geq\frac{4}{3}.$$
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, the condition does not depend on $v^2$ and we need to prove that
$$9v^4-6uw^3\geq\frac{4}{3}w^6,$$
which says that it's enough to prove the last inequality for a minimal value of $v^2$.
We know that $a$, $b$ and $c$ are positive roots of the equation
$$(x-a)(x-b)(x-c)=0$$ or
$$x^3-3ux^2+3v^2x-w^3=0$$ or
$$3v^2x=-x^3+3ux^2+w^3.$$
Thus, the line $y=3v^2x$ and the graph of $y=-x^3+3ux^2+w^3$ have three common points 
and $v^2$ gets a minimal value, 
when the line $y=3v^2x$ is a tangent line to the graph of $y=-x^3+3ux^2+w^3$,
which happens for equality case of two variables (draw it!).
Id est, it's enough to prove the last inequality for $b=a$ and the condition gives $c=\frac{27+36a}{32a^2-18}$.
Thus, we need to prove that:
$$a^4+2a^2\left(\frac{27+36a}{32a^2-18}\right)^2\geq\frac{4}{3}a^4\left(\frac{27+36a}{32a^2-18}\right)^2$$ or
$$a^2(2a-3)^2(8a^2+12a+9)\geq0.$$ 
Done!
