<s>[INST] If $a,b,c \geq 1$ then prove $\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1} < \sqrt{abc+c}$ If $a,b,c \geq 1$, then prove:
$$\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1}  <  \sqrt{abc+c}$$
My try:
$$A=\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1} \\ A^2 =(a-1)+(b-1)+(c-1)+2\sqrt{(a-1)(b-1)}+2\sqrt{(a-1)(c-1)}+2\sqrt{(b-1)(c-1)} $$
$$B=\sqrt{abc+c} \\B^2=abc+c=c(ab+1) $$
Now what?
 [/INST] The given inequality is equivalent to 
$$ \sqrt{a-1} + \sqrt{b-1} \leq \sqrt{abc + c} - \sqrt{c-1}.$$
Now set $f(c) := \gamma \sqrt{c} - \sqrt{c-1}$ where $ \gamma := \sqrt{ab+1} $.  Then we see that $f'(c) = \gamma \frac{1}{2\sqrt{c}} - \frac{1}{2\sqrt{c-1}}$, from which it is easy to check $f$ takes its minimum when $c=\frac{1}{\gamma^2-1} + 1 = \frac{1}{ab} + 1$. So what we have to prove is 
$$ \sqrt{a-1} + \sqrt{b-1} \leq f(\frac{1}{ab} + 1) = \sqrt{ab}$$
which is equivalent to 
$$ a-1 + b-1 + 2 \sqrt{(a-1)(b-1)} \leq ab$$
or, $2\sqrt{(a-1)(b-1)} \leq (a-1)(b-1) + 1$. The last inequality can be easily derived from AM-GM applied to $(a-1)(b-1)$ and $1$. In addition we see that the equality holds iff $(a-1)(b-1) = 1$ and $c = \frac{1}{ab} + 1$.
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