\(Q=\sqrt{2a+bc}+\sqrt{2b+ca}+\sqrt{2c+ab}\)
\(\Rightarrow Q^2=\left(\sqrt{2a+bc}+\sqrt{2b+ca}+\sqrt{2c+ab}\right)^2\)
Vì \(a,b,c>0\)nên áp dụng bất đẳng thức Bunhiacopxki, ta được:
\(\left(\sqrt{2a+bc}+\sqrt{2b+ca}+\sqrt{2c+ab}\right)^2\)\(\le\left(1^2+1^2+1^2\right)\left[\left(\sqrt{2a+bc}\right)^2+\left(\sqrt{2b+ca}\right)^2+\left(\sqrt{2c+ab}\right)^2\right]\)
\(\Leftrightarrow Q^2\le3\left(2a+bc+2b+ca+2c+ab\right)\)
\(\Leftrightarrow Q^2\le3\left[2\left(a+b+c\right)+\left(ab+bc+ca\right)\right]\)
\(\Leftrightarrow Q^2\le6\left(a+b+c\right)+3\left(ab+bc+ca\right)\)
\(\Leftrightarrow Q^2\le6.2+3\left(ab+bc+ca\right)\)(vì \(a+b+c=2\))
\(\Leftrightarrow Q^2\le12+3\left(ab+bc+ca\right)\left(1\right)\)
Vì \(a,b,c>0\)nên áp dụng bất dẳng thức Cô-si cho 2 số dương, ta được:
\(a^2+b^2\ge2ab\left(2\right)\);
\(b^2+c^2\ge2bc\left(3\right)\)
\(c^2+a^2\ge2ca\left(4\right)\)
Từ \(\left(2\right),\left(3\right),\left(4\right)\), ta được:
\(a^2+b^2+b^2+c^2+c^2+a^2\ge2ab+2bc+2ca\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca\ge\)\(ab+bc+ca+2ab+2bc+2ca\)
\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Leftrightarrow2^2\ge3\left(ab+bc+ca\right)\)(vì \(a+b+c=2\))
\(\Leftrightarrow4\ge3\left(ab+bc+ca\right)\)
\(\Leftrightarrow4+12\ge3\left(ab+bc+ca\right)+12\)
\(\Leftrightarrow3\left(ab+bc+ca\right)+12\le16\left(5\right)\)
Từ (1) và (5), ta được:
\(Q^2\le16\)
\(\Leftrightarrow Q\le4\)
Dấu bằng xảy ra.
\(\Leftrightarrow\hept{\begin{cases}a=b=c>0\\a+b+c=2\end{cases}}\Leftrightarrow a=b=c=\frac{2}{3}\)
Vậy \(maxQ=4\Leftrightarrow a=b=c=\frac{2}{3}\)