Với mọi số thực dương x;y;z ta luôn có:
\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Leftrightarrow2x^2+2y^2+2z^2\ge2xy+2yz+2xz\)
\(\Leftrightarrow3\left(x^2+y^2+z^2\right)\ge x^2+y^2+z^2+2xy+2yz+2zx\)
\(\Leftrightarrow3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)
\(\Leftrightarrow x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}\)
Áp dụng:
\(P\le\sqrt{3\left(3a+b+3b+c+3c+a\right)}=\sqrt{12\left(a+b+c\right)}=\sqrt{36}=6\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Sửa: \(P=\sqrt{3a+bc}+\sqrt{3b+ca}+\sqrt{3c+ab}\)
Áp dụng BĐT cosi: \(x+y\ge2\sqrt{xy}\Leftrightarrow\sqrt{xy}\le\dfrac{x+y}{2}\)
Ta có \(\sqrt{3a+bc}=\sqrt{\left(a+b+c\right)a+bc}=\sqrt{a^2+ab+ac+bc}\)
\(=\sqrt{\left(a+b\right)\left(a+c\right)}\le\dfrac{a+b+a+c}{2}=\dfrac{2a+b+c}{2}\)
Cmtt \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{3b+ca}\le\dfrac{a+2b+c}{2}\\\sqrt{3c+ab}\le\dfrac{a+b+2c}{2}\end{matrix}\right.\)
Cộng VTV 3 BĐT trên, ta được:
\(P\le\dfrac{2a+b+c+a+2b+c+a+b+2c}{2}\\ \Leftrightarrow P\le\dfrac{4\left(a+b+c\right)}{2}=2\left(a+b+c\right)=2\cdot3=6\)
Vậy \(P_{max}=6\Leftrightarrow a=b=c=1\)
