Áp dụng bất đẳng thức Cô-si và Bunhiacopxki cho ba số không âm a,b,c, ta có:
- \(a^2+b^2+c^2\le\dfrac{\left(a+b+c\right)^2}{3}\)
- \(abc\le\dfrac{\left(a+b+c\right)^3}{27}\)
\(\Rightarrow P\ge\dfrac{1}{\dfrac{\left(a+b+c\right)^2}{3}}+\dfrac{1}{\dfrac{\left(a+b+c\right)^3}{27}}=3+27=30\)
Vậy GTNN của P = 30 khi a = b = c = 1/3
☘ \(abc\le\dfrac{1}{9}\left(ab+bc+ca\right)\left(a+b+c\right)=\dfrac{ab+bc+ca}{9}\)
☘ \(ab+bc+ca\le\dfrac{1}{3}\left(a+b+c\right)^2=\dfrac{1}{3}\)
\(VT\ge\dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ca}=\dfrac{1}{a^2+b^2+c^2}+\dfrac{4}{2\left(ab+bc+ca\right)}+\dfrac{7}{ab+bc+ca}\)
\(\ge\dfrac{\left(1+2\right)^2}{\left(a+b+c\right)^2}+\dfrac{7}{\dfrac{1}{3}}=9+21=30\)
