Áp dụng bất đẳng thức Cô – si ta có:

$2.\frac{4}{3}\sqrt{2\text{a}+bc}\le {\left(\frac{4}{3}\right)}^2+2\text{a}+bc$

$2.\frac{4}{3}\sqrt{2b+ca}\le {\left(\frac{4}{3}\right)}^2+2b+ca$

$2.\frac{4}{3}\sqrt{2c+ab}\le {\left(\frac{4}{3}\right)}^2+2c+ab$

Suy ra

$2.\frac{4}{3}\sqrt{2\text{a}+bc}+2.\frac{4}{3}\sqrt{2b+ca}+2.\frac{4}{3}\sqrt{2c+ab}\le {\left(\frac{4}{3}\right)}^2+2\text{a}+bc+{\left(\frac{4}{3}\right)}^2+2b+ac+{\left(\frac{4}{3}\right)}^2+2c+ab$

$\iff \frac{8}{3}\left(\sqrt{2\text{a}+bc}+\sqrt{2b+ca}+\sqrt{2c+ab}\right)\le \frac{16}{3}+2\text{a}+2b+2c+ab+ac+bc$

$\iff \frac{8}{3}Q\le \frac{16}{3}+2\left(a+b+c\right)+ab+ac+bc$

$\iff \frac{8}{3}Q\le \frac{16}{3}+4+ab+ac+bc$

$\iff \frac{8}{3}Q\le \frac{28}{3}+ab+ac+bc$

Ta có

3(ab + bc + ca)

= 2(ab + bc + ca) + ab + bc + ca

≤ 2(ab + bc + ca) + a² + b² + c² = (a + b + c)² = 2² = 4

Suy ra $ab+bc+ca\le \frac{4}{3}$

Do đó $\frac{8}{3}Q\le \frac{28}{3}+\frac{4}{3}$

Hay Q ≤ 4

Dấu “ = ” xảy ra khi $\text{a}=b=c=\frac{2}{3}$

Vậy $\text{a}=b=c=\frac{2}{3}$  thì Q đạt giá trị lớn nhất bằng 4.