Có \(\sqrt{2a+bc}\)=\(\sqrt{\left(a+b+c\right)a+bc}\)=\(\sqrt{a^2+ab+ac+bc}\)=\(\sqrt{\left(a+c\right)\left(a+b\right)}\)
AD bđt cosi vs hai số dương có:
\(\sqrt{\left(a+c\right)\left(a+b\right)}\le\frac{a+c+a+b}{2}\) hay \(\sqrt{2a+bc}\le\frac{2a+b+c}{2}\) (1)
CM tương tự cx có: \(\sqrt{2b+ac}\le\frac{2b+a+c}{2}\) (2)
\(\sqrt{2c+ab}\le\frac{2c+a+b}{2}\) (3)
Từ (1),(2),(3) => Q=\(\sqrt{2a+bc}+\sqrt{2b+ac}+\sqrt{2c+ab}\le\frac{2a+b+c+2b+a+c+2c+a+b}{2}\)
<=> Q \(\le\frac{4\left(a+b+c\right)}{2}=\frac{4.2}{2}\)(vì a+b+c=2)
<=> Q\(\le4\)
Dấu "=" xảy ra <=> a=b=c=\(\frac{2}{3}\)
vậy GTLN của Q=4 <=> a=b=c=\(\frac{2}{3}\)
