dự đoán dấu đẳng thức xảy ra tại \(a=b=c=\sqrt{3}-1\) nên \(MinM=\dfrac{3\sqrt{3}}{4}\)ta đi chứng minh \(M\le\dfrac{3\sqrt{3}}{4}\)
Đặt \(a+1=x\); \(b+1=y\); \(c+1=z\)
\(\Rightarrow x+y+z=a+b+c+3=a+b+c+ab+bc+ac+abc+1\)
\(=\left(a+1\right)\left(b+1\right)\left(c+1\right)\)\(=xyz\)
\(\Rightarrow x+y+z=xyz\) đặt \(x=tan\dfrac{A}{2};y=tan\dfrac{B}{2};z=tan\dfrac{C}{2}\)
\(\Rightarrow tan\dfrac{A}{2}+tan\dfrac{B}{2}+tan\dfrac{C}{2}=tan\dfrac{A}{2}.tan\dfrac{B}{2}.tan\dfrac{C}{2}\)(đúng , tự cm hoặc google)
do \(x=tan\dfrac{A}{2}\Rightarrow sinA=\dfrac{2x}{x^2+1}\Rightarrow\dfrac{sinA}{2}=\dfrac{x}{x^2+1}\)
\(\Rightarrow M=\Sigma\dfrac{sinA}{2}\)ta phai cm \(M=\Sigma\dfrac{sinA}{2}\le\dfrac{3\sqrt{3}}{4}\Leftrightarrow M\le\dfrac{3\sqrt{3}}{2}\)
\(\Leftrightarrow sinA+sinB+sinC\le\dfrac{3\sqrt{3}}{2}\)theo BDT Cauchy-Swarch
\(\Rightarrow sinA+sinB+sinC\le\sqrt{3\left(sinA^2+sin^2B+sin^2C\right)}\)
mặt khác \(sin^2A+sin^2B+sin^2C\le\dfrac{9}{4}\)(trong sach toan 10 co BDT nay hoac google)
\(\Rightarrow sinA+sinB+sinC\le\sqrt{\dfrac{3.9}{4}}=\dfrac{3\sqrt{3}}{2}\)
\(\Rightarrow dpcm\)
