Trước tiên chứng minh BĐT \(\frac{x^3+1}{x+2}\ge\frac{7}{18}x^2+\frac{5}{18}\left(x>0\right)\)
\(\Leftrightarrow18\left(x^3+1\right)\ge\left(x+2\right)\left(7x^2+5\right)\)
\(\Leftrightarrow\left(x-1\right)^2\left(11x+8\right)\ge0\)(luôn đúng với x>0)
Dấu "=" xảy ra khi x = 1
Áp dụng công thức trên ta có:
Cho x lần lượt là \(\frac{a}{b};\frac{b}{c};\frac{c}{a}\)
\(\Leftrightarrow\frac{a^3+b^3}{a+2b}\ge\frac{7a^2}{18}+\frac{5b^2}{18};\frac{b^3+c^3}{a+2b}\ge\frac{7b^2}{18}+\frac{5c^2}{18};\frac{c^3+a^3}{a+2b}\ge\frac{7c^2}{18}+\frac{5a^2}{18}\)
Từ đẳng thức trên suy ra \(A\ge\frac{12+\left(a^2+b^2+c^2\right)}{18}=2\)
Vậy MinA=2 khi a=b=c=1
Cần cm: \(\frac{a^3+b^3}{a+2b}\ge\frac{7}{18}a^2+\frac{5}{18}b^2\)
bđt \(\Leftrightarrow\)\(11a^3+8b^3-14a^2b-5ab^2\ge0\)\(\Leftrightarrow\)\(\left(a-b\right)^2\left(11a+8b\right)\ge0\) đúng với a,b>0
\(A\ge\frac{2}{3}\left(a^2+b^2+c^2\right)=2\)
Dấu "=" xảy ra khi a=b=c=1
Cach khac
Ta co:
\(A=M_1+M_2\)
Voi \(\hept{\begin{cases}M_1=\frac{a^3}{a+2b}+\frac{b^3}{b+2c}+\frac{c^3}{c+2a}\\M_2=\frac{a^3}{c+2a}+\frac{b^3}{a+2b}+\frac{c^3}{b+2c}\end{cases}}\)
Xet
\(M_1=\Sigma_{cyc}\frac{a^3}{a+2b}\ge\frac{\left(a^2+b^2+c^2\right)^2}{\Sigma_{cyc}a^2+2\Sigma_{cyc}ab}=\frac{9}{\left(a+b+c\right)^2}\ge\frac{9}{3\left(a^2+b^2+c^2\right)}=1\)
\(M_2=\Sigma_{cyc}\frac{a^3}{ca+2a^2}\ge\frac{\left(a^2+b^2+c^2\right)^2}{\Sigma_{cyc}ab+2\Sigma_{cyc}a^2}\ge\frac{9}{ab+bc+ca+6}\ge\frac{9}{a^2+b^2+c^2+6}=1\)
\(\Rightarrow A\ge2\)
Dau '=' xay ra khi \(a=b=c=1\)
