Theo Cô-si: \(\dfrac{a^3}{b}+ab\ge2\sqrt{\dfrac{a^4b}{b}}=2a^2\)
Tương tự có: \(\dfrac{b^3}{c}+bc\ge2b^2\) ; \(\dfrac{c^3}{a}+ac\ge2c^2\)
\(\Rightarrow\) P + ab+bc+ca \(\ge\) 2 (a2+b2+c2)
Mà a2+b2+c2 \(\ge\) ab+bc+ca
\(\Rightarrow\) P+ ab+bc+ca \(\ge\) a2+b2+c2 +ab+bc+ca
\(\Leftrightarrow\) P \(\ge\) a2+b2+c2
\(\Leftrightarrow\) 3P \(\ge\) 2( a2+b2+c2)+( a2+b2+c2)
Có: a2+b2+c2 \(\ge\)ab+bc+ca
\(\Rightarrow\)3P\(\ge\) 2(ab+bc+ca) + a2+ 1 +b2 +1+ c2 +1 -3
Lại có: a2+1\(\ge\) 2a ; b2+1\(\ge\) 2b ; c2+a\(\ge\) 2c
\(\Rightarrow\) 3P \(\ge\) 2(ab+bc+ca) +2a+2b+2c - 3
\(\Leftrightarrow\)3P\(\ge\) 2(ab+bc+ca +a+b+c) -3 = 2.6-3=9
\(\Leftrightarrow\)P\(\ge\)3
Vậy Pmin = 3
Dấu "=" xảy ra\(\Leftrightarrow\) a=b=c=1
