\(\text{Áp dụng BĐT Bunhiacopxki ta có:}\)
\(\left(\sqrt{a-1}+\sqrt{b-1}\right)^2=\left(1\sqrt{a-1}+1\sqrt{b-1}\right)^2\le\left(a-1+1\right)\left(b+1-1\right)=ab\)
\(\text{Dễ chứng minh được:}\) \(\sqrt{ab}+\sqrt{c-1}\le\sqrt{c\left(a+1\right)}\)
\(\text{Áp dụng BĐT Bunhiacopxki ta có:}\)
\(\left(1\sqrt{ab}+1\sqrt{c-1}\right)^2\le\left(ab+1\right)\left(1+c-1\right)=c\left(ab+1\right)\)
\(\Leftrightarrow\sqrt{ab}+\sqrt{c-1}\le\sqrt{c\left(ab+1\right)}\)
Từ trên ta có: \(\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\le\sqrt{ab}+\sqrt{c-1}\le\sqrt{c\left(ab+1\right)}\)
\(\Rightarrow\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\le\sqrt{c\left(ab+1\right)}\left(đpcm\right)\)
