Ta có:
\(\frac{a^2}{b}+9a^2b\ge2\sqrt{9a^4}=6a^2\)
Suy ra \(\frac{a^2}{b}\ge6a^2-9a^2b\)
Tương tự hai BĐT còn lại rồi cộng theo vế suy ra
\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge6\left(a^2+b^2+c^2\right)-9\left(a^2b+b^2c+c^2a\right)\) (*)
Mặt khác ta có BĐT sau: \(\left(a^2+b^2+c^2\right)\left(a+b+c\right)\ge3\left(a^2b+b^2c+c^2a\right)\)
\(\Leftrightarrow a\left(a-b\right)^2+b\left(b-c\right)^2+c\left(c-a\right)^2\ge0\) (đúng)
Do đó \(\left(a^2+b^2+c^2\right)=\left(a^2+b^2+c^2\right)\left(a+b+c\right)\ge3\left(a^2b+b^2c+c^2a\right)\)
Thay vào (*) ta có: \(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge6\left(a^2+b^2+c^2\right)-9\left(a^2b+b^2c+c^2a\right)\ge3\left(a^2+b^2+c^2\right)\)
Thay vào P: \(P=2018\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)+\frac{1}{3\left(a^2+b^2+c^2\right)}\)
\(\ge2018.3\left(a^2+b^2+c^2\right)+\frac{1}{3\left(a^2+b^2+c^2\right)}\)
\(=2017.3\left(a^2+b^2+c^2\right)+3\left(a^2+b^2+c^2\right)+\frac{1}{3\left(a^2+b^2+c^2\right)}\)
\(\ge2017\left(a+b+c\right)^2+2=2019\)
Đẳng thức xảy ra khi a = b = c= 1/3
P/s: Em trình bày hơi lủng củng nha!
Chợt nghĩ ra cách khác:Chú ý BĐT: \(\left(a^2+b^2+c^2\right)\left(a+b+c\right)\ge3\left(a^2b+b^2c+c^2a\right)\)
Có:\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}=\frac{a^4}{a^2b}+\frac{b^4}{b^2c}+\frac{c^4}{c^2a}\)
\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^2b+b^2c+c^2a}\ge\frac{3\left(a^2+b^2+c^2\right)^2}{a^2+b^2+c^2}=3\left(a^2+b^2+c^2\right)\)
Rồi đến đây ok:v
