\(\dfrac{\sqrt{a^2-1}}{a}=\dfrac{\sqrt{3\left(a^2-1\right)}}{\sqrt{3}a}\le\dfrac{\dfrac{a^2+2}{2}}{a\sqrt{3}}=\dfrac{a^2+2}{2a\sqrt{3}}\)
\(\dfrac{\sqrt{b^2-1}}{b}\le\dfrac{b^2+2}{2b\sqrt{3}}\)
\(\Rightarrow F\le\dfrac{a^2+2}{2a\sqrt{3}}+\dfrac{b^2+2}{2b\sqrt{3}}+\dfrac{1}{2ab}=\dfrac{2ab-4}{2\sqrt{3}}+\dfrac{1}{a\sqrt{3}}+\dfrac{1}{b\sqrt{3}}+\dfrac{1}{2ab}=\dfrac{ab}{\sqrt{3}}+\dfrac{1}{a\sqrt{3}}+\dfrac{1}{b\sqrt{3}}+\dfrac{1}{2ab}-\dfrac{4}{2\sqrt{3}}=\dfrac{2a^2b^2+2\left(a+b\right)+\sqrt{3}}{2ab\sqrt{3}}-\dfrac{4}{2\sqrt{3}}=\dfrac{2\left(ab\right)^2+2\left(2ab-4\right)+\sqrt{3}}{2ab\sqrt{3}}\)
có:: \(2ab=a+b+4\ge2\sqrt{ab}+4\Rightarrow2ab-2\sqrt{ab}-4\ge0\)
\(\Rightarrow2\sqrt{ab}^2-2\sqrt{ab}-4\ge0\Rightarrow\sqrt{ab}\ge2\Rightarrow ab\ge4\)
\(đặt:ab=t\ge4\)
\(\Rightarrow F\le\dfrac{2t^2+2\left(2t-4\right)+\sqrt{3}}{2t\sqrt{3}}-\dfrac{4}{2\sqrt{3}}=\dfrac{2t^2+4t-8+\sqrt{3}}{2t\sqrt{3}}-\dfrac{4}{2\sqrt{3}}\le\dfrac{2t^2+4t-8+\sqrt{4}}{8\sqrt{3}}-\dfrac{4}{2\sqrt{3}}\)
\(xét\) \(f\left(t\right)=2t^2+4t-8+\sqrt{3}\left(t\ge4\right)\Rightarrow maxf\left(t\right)=40+\sqrt{3}\)
\(\Rightarrow F\le\dfrac{40+\sqrt{3}}{8\sqrt{3}}-\dfrac{4}{2\sqrt{3}}=\dfrac{1+8\sqrt{3}}{8}\)
\(dâu"="\Leftrightarrow a=b=2\)
