Sử dụng bđt bunyakovsky 2 lần : $(1.\sqrt{b-1}+\sqrt{a-1}.1)^{2}\leq (1+a-1)(1+b-1)\Rightarrow \sqrt{a-1}+\sqrt{b-1}\leq \sqrt{ab}$ $(\sqrt{ab}.1+1.\sqrt{c-1})^{2}\leq (ab+1)(c-1+1)\Rightarrow \sqrt{ab}+\sqrt{c-1}\leq \sqrt{c(ab+1)}$
Ta có VT$=\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\leq \sqrt{ab}+\sqrt{c-1}\leq \sqrt{c(ab+1)}\Rightarrow $(đpcm)