Ta có : \(P=\frac{bc}{\sqrt{3a+bc}}+\frac{ca}{\sqrt{3b+ca}}+\frac{ab}{\sqrt{3c+ab}}\)
\(=\frac{bc}{\sqrt{a^2+ab+ac+bc}}+\frac{ca}{\sqrt{ab+b^2+bc+ac}}+\frac{ab}{\sqrt{ac+bc+c^2+ab}}\)
\(=\frac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{ca}{\sqrt{\left(b+a\right)\left(b+c\right)}}+\frac{ab}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)
Áp dụng bdt Cauchy : \(\frac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{bc}{2}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\) ; \(\frac{ac}{\sqrt{\left(b+c\right)\left(b+a\right)}}\le\frac{ac}{2}\left(\frac{1}{b+c}+\frac{1}{b+a}\right)\)
\(\frac{ab}{\sqrt{\left(c+a\right)\left(c+b\right)}}\le\frac{ab}{2}\left(\frac{1}{c+a}+\frac{1}{c+b}\right)\)
Suy ra : \(A\le\frac{1}{2}\left(\frac{bc}{a+b}+\frac{bc}{a+c}+\frac{ac}{b+c}+\frac{ac}{a+b}+\frac{ab}{c+a}+\frac{ab}{b+c}\right)\)
\(\Rightarrow A\le\frac{1}{2}\left(\frac{bc+ac}{a+b}+\frac{bc+ab}{a+c}+\frac{ac+ab}{b+c}\right)=\frac{1}{2}\left(a+b+c\right)=\frac{3}{2}\)
\(A\le\frac{3}{2}\Rightarrow MaxA=\frac{3}{2}\Leftrightarrow a=b=c=1\)
\(\frac{ab}{\sqrt{3c+ab}}\le\frac{ab}{\sqrt{ab+\left(a+b+c\right)c}}=\frac{ab}{\sqrt{\left(b+c\right)\left(c+a\right)}}=\frac{\sqrt{ab}}{\sqrt{b+c}}\cdot\frac{\sqrt{ab}}{\sqrt{c+a}}\)\(=\sqrt{\frac{ab}{b+c}\cdot\frac{ab}{c+a}}\le\frac{1}{4}\left(\frac{2ab}{b+c}+\frac{2ab}{c+a}\right)\)
Tương tự cho 2 cái kia rồi cộng lại
\(P\le\frac{1}{4}\left(2a+2b+2c\right)\le\frac{3}{2}\)
\(Max_P=\frac{3}{2}\Leftrightarrow a=b=c=1\)
chưa học cosi thì tui cho mấy cách thik cách nào thì xìa
Giúp mk vs m.n
