<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] 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.
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