<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] $$\sqrt{a-1}+\sqrt{b-1}=\sqrt{a+b-2+2\sqrt{(a-1)(b-1)}}=$$
$$=\sqrt{ab-(a-1)(b-1)+2\sqrt{(a-1)(b-1)}-1}=$$
$$=\sqrt{ab-\left(\sqrt{(a-1)(b-1)}-1\right)^2}\leq\sqrt{ab}.$$
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!
</s>