\(\frac{1}{\sqrt[3]{x+3y}}\ge\frac{1}{\frac{x+3y+1+1}{3}}=\frac{3}{x+3y+2}\\ \text{Tương tự }\Rightarrow P\ge\frac{3}{x+3y+2}+\frac{3}{y+3z+2}+\frac{3}{z+3x+2}\\ \ge3\cdot\frac{9}{x+3y+2+y+3z+2+z+3x+2}\\ =3\)
Ta có: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca\ge0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\ge0\)(với a,b,c > 0 )
\(\Leftrightarrow a^3+b^3+c^3\ge3abc\Leftrightarrow abc\le\frac{a^3+b^3+c^3}{3}\).
AD CT trên ta có :
\(1.1.\sqrt[3]{x+3y}\le\frac{1+1+x+3y}{3}\Leftrightarrow\sqrt[3]{x+3y}\le\frac{x+3y+2}{3}\).
Cmtt có : \(\sqrt[3]{y+3z}\le\frac{y+3z+2}{3};\sqrt[3]{z+3x}\le\frac{z+3x+2}{3}\)
\(\Rightarrow\sqrt[3]{x+3y}+\sqrt[3]{y+3z}+\sqrt[3]{z+3x}\le\frac{4\left(x+y+z\right)+6}{3}=3\)
AD BĐT Cộng mẫu số ta có:
\(\frac{1}{\sqrt[3]{x+3y}}+\frac{1}{\sqrt[3]{y+3z}}+\frac{1}{\sqrt[3]{z+3x}}\ge\frac{\left(1+1+1\right)^2}{\sqrt[3]{x+3y}+\sqrt[3]{y+3z}+\sqrt[3]{z+3x}}\ge\frac{9}{3}=3\)Dấu ''='' xảy ra \(\Leftrightarrow a=b=c=\frac{1}{4}\)
Vậy GTNN của b.thức là P = 3 khi a = b = c =\(\frac{1}{4}\)
