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?
 A: 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 $c = \frac{1}{ab} + 1$.
A: Substitute $a=x^2+1$, $b=y^2+1$ and $c=z^2+1$ for $x,y,z\ge 0$ to get
$$x+y+z\le\sqrt{(z^2+1)\left(\left(x^2+1\right)\left(y^2+1\right)+1\right)}$$
Now after squaring and grouping we obtain
\begin{align*}x^2+y^2+z^2+2xy+2yz+2zx\le(z^2+1)\left((x^2+1)(y^2+1)+1\right) \\
0\le z^2(x^2+1)(y^2+1)-2z(x+y)+x^2y^2-2xy+2 \\
0\le z^2(x^2+1)(y^2+1)-2z(x+y)+(xy-1)^2+1
\end{align*}
We'll prove that $z^2(x^2+1)(y^2+1)-2z(x+y)+1\ge 0$. The determinant of this quadratic in $z$ is $\Delta =4(x+y)^2-4(x^2+1)(y^2+1)=-4(xy-1)^2\le0$, hence the inequality is true.
A: $$\sqrt{a-1}+\sqrt{b-1}=\sqrt{a+b-2+2\sqrt{(a-1)(b-1)}}=$$
$$=\sqrt{ab-(a-1)(b-1)+2\sqrt{(a-1)(b-1)}-1}=$$
$$=\sqrt{ab-\left(\sqrt{(a-1)(b-1)}-1\right)^2}\leq\sqrt{ab}.$$
Thus,
$$\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\leq\sqrt{ab}+\sqrt{c-1}=$$
$$=\sqrt{ab+1-1}+\sqrt{c-1}\leq\sqrt{(ab+1)c}=\sqrt{abc+c}.$$
Done!
