\(\sqrt{9+tan^4A}\ge\sqrt{2\sqrt{9.tan^4A}}=\sqrt{6}.tanA\) , chứng minh tương tự
\(\Rightarrow\sqrt{9+tan^4A}+\sqrt{9+tan^4B}+\sqrt{9+tan^4C}\ge\sqrt{6}\left(tanA+tanB+tanC\right)\)
lại có trong tam giác ABC:
\(A+B+C=180^0\Rightarrow A+B=180^0-C\Rightarrow tan\left(A+B\right)=tan\left(180^0-C\right)\)
\(\Rightarrow\dfrac{tanA+tanB}{1-tanA.tanB}=tan\left(180^0-C\right)=-tanC\)
\(\Rightarrow tanA+tanB=-tanC+tanA.tanB.tanC\)
\(\Rightarrow tanA+tanB+tanC=tanA.tanB.tanC\)
Do ABC là tam giác nhọn \(\Rightarrow tanA,tanB,tanC>0\)
Áp dụng BĐT Cauchy: \(tanA+tanB+tanC\ge3\sqrt[3]{tanA.tanB.tanC}\)
\(\Rightarrow\dfrac{\left(tanA+tanB+tanC\right)^3}{27}\ge tanA.tanB.tanC=tanA+tanB+tanC\)
\(\Rightarrow\left(tanA+tanB+tanC\right)^2\ge27\) \(\Rightarrow tanA+tanB+tanC\ge3\sqrt{3}\)
\(\Rightarrow\sqrt{6}\left(tanA+tanB+tanC\right)\ge\sqrt{6}.3\sqrt{3}=9\sqrt{2}\)
\(\Rightarrow\sqrt{9+tan^4A}+\sqrt{9+tan^4B}+\sqrt{9+tan^4C}\ge9\sqrt{2}\)
Dấu "=" xảy ra khi tam giác ABC đều
