Đặt A = \(\sqrt{2a+bc}+\sqrt{2b+ca}+\sqrt{2c+ab}\)
Áp dụng BĐT bu-nhi-a-cốp-xki ta có:
\(A^2=\left(\sqrt{2a+bc}+\sqrt{2b+ac}+\sqrt{2c+ab}\right)^2\le\left(1^2+1^2+1^2\right)\left(2a+bc+2b+ac+2c+ab\right)\)Mà ta lại có: \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\) (bn tự cm)
\(\Rightarrow\dfrac{\left(a+b+c\right)^2}{3}\ge\left(ab+bc+ca\right)\Leftrightarrow ab+bc+ca\le\dfrac{4}{3}\)
Từ đó \(A^2\le\left(1^2+1^2+1^2\right)\left(2.2+\dfrac{4}{3}\right)=16\)
Hay \(A\le4\). Đẳng thức xảy ra \(\Leftrightarrow a=b=c=\dfrac{2}{3}\)
\(S=\sqrt{2a+bc}+\sqrt{2c+ab}+\sqrt{2b+ac}\\ \Rightarrow\dfrac{32}{9}S=2\sqrt{2a+bc}\cdot\dfrac{16}{9}+2\sqrt{2c+ab}\cdot\dfrac{16}{9}+2\sqrt{2b+ac}\cdot\dfrac{16}{9}\\ \le2a+bc+\dfrac{16}{9}+2c+ab+\dfrac{16}{9}+2b+ac+\dfrac{16}{9}\\ =ab+ac+bc+2a+2b+2c+\dfrac{16}{3}\\ \le\dfrac{\left(a+b+c\right)^2}{3}+2\left(a+b+c\right)+\dfrac{16}{3}\\ =\dfrac{4}{3}+4+\dfrac{16}{3}=\dfrac{32}{3}\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}a=b=c\\a+b+c=2\end{matrix}\right.\Leftrightarrow a=b=c=\dfrac{2}{3}\)
Vậy \(S_{Max}=\dfrac{32}{3}\) khi \(a=b=c=\dfrac{2}{3}\)
\(S=\sqrt{2a+bc}+\sqrt{2c+ab}+\sqrt{2b+ac}\\ \Rightarrow\dfrac{8}{3}S=2\sqrt{2a+bc}\cdot\dfrac{4}{3}+2\sqrt{2c+ab}\cdot\dfrac{4}{3}+2\sqrt{2b+ac}\cdot\dfrac{4}{3}\\ \ge2a+bc+\dfrac{16}{9}+2c+ab+\dfrac{16}{9}+2b+ac+\dfrac{16}{9}\)
\(\Rightarrow S\le4\)
Vậy \(S_{Max}=4\) khi \(a=b=c=\dfrac{2}{3}\)
