thôi gợi ý :v dg chán đời lại mới coi xong anime, "Happy Ending" :v
By AM-GM have:
\(xy+xz+yz \ge 3\sqrt[3]{x^2y^2z^2} \)
\(\sqrt[3]{xyz} \le \sqrt{\dfrac{x^2+y^2+z^2}{3}} \)
\(x+y+z \le \sqrt{3(x^2+y^2+z^2)}\)
Another way
BĐT trên thuần nhất nên ta chuẩn hóa \(x^2+y^2+z^2=1\)
\(BDT\Leftrightarrow\dfrac{xyz\left(x+y+z+1\right)}{xy+yz+xz}\le\dfrac{3+\sqrt{3}}{9}\)
Áp dụng BĐT AM-GM ta có:
\(\left\{{}\begin{matrix}xy+yz+xz\ge3\sqrt{x^2y^2z^2}\\\sqrt[3]{xyz}\le\sqrt{\dfrac{x^2+y^2+z^2}{3}}\Rightarrow xyz\le\sqrt{\dfrac{\left(x^2+y^2+z^2\right)^3}{27}}=\dfrac{1}{\sqrt{27}}\\x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow A\le\dfrac{\sqrt[3]{xyz}\left(x+y+z+\sqrt{x^2+y^2+z^2}\right)}{3\left(x^2+y^2+z^2\right)}\)
\(=\dfrac{\sqrt[3]{xyz}\left(x+y+z+1\right)}{3}\le\dfrac{3+\sqrt{3}}{9}\)
