\(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\ge2+\sqrt{22+\frac{1}{abc}}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{ac}}+\frac{1}{\sqrt{bc}}\right)\)
\(\ge2+\sqrt{22+\frac{\left(a+b+c\right)^3}{abc}}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)
\(\ge2\sqrt{abc}+\sqrt{22abc+\left(a+b+c\right)^3}\)
\(\Leftrightarrow\left(a+b+c\right)^2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\)
\(\ge\left(a+b+c\right)^3+26abc+4\sqrt{abc}\sqrt{\left(a+b+c\right)^3+22abc}\)
\(\Leftrightarrow\left(a+b+c\right)^2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)
\(\ge13abc+2\sqrt{abc}\sqrt{\left(a+b+c\right)^3+22abc}\left(1\right)\)
Ta có:
\(VT_{\left(1\right)}=\frac{13}{27}\left(a+b+c\right)^2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+\frac{14}{27}\left(a+b+c\right)^2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)
\(\ge13abc+\frac{14}{27}\left(a+b+c\right)^2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)
Ta cần chứng minh:
\(\frac{14}{27}\left(a+b+c\right)^2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)
\(\ge2\sqrt{abc}\sqrt{\left(a+b+c\right)^3+22abc}\)
\(\Leftrightarrow7\left(a+b+c\right)^2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)
\(\ge27\sqrt{abc}\sqrt{\left(a+b+c\right)^3+22abc}\)
\(\Leftrightarrow49\left(a+b+c\right)^4\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)^2\)
\(\ge27^2abc\left[\left(a+b+c\right)^3+22abc\right]\left(2\right)\)
Lại có:
\(VT_{\left(2\right)}\ge49\left(a+b+c\right)^3\cdot27abc\)
Ta chứng minh
\(49\left(a+b+c\right)^3\cdot27abc\ge27^2abc\left[\left(a+b+c\right)^3+22abc\right]\)
\(\Leftrightarrow\left(a+b+c\right)^3\ge27abc\)
Bất đẳng thức cuối cùng hiển nhiên đúng theo Cô-si 3 số.
Vậy bđt đã được chứng minh.
Dấu = khi a=b=c
