dễ cm \(\frac{9\left(a+b\right)\left(b+c\right)\left(c+a\right)}{4\left(ab+bc+ca\right)}\ge2\left(a+b+c\right)\)
\(2\sqrt{a^2-ab+b^2}=2\sqrt{\left(\frac{a^2}{b}-a+b\right)b}\le a^2-a+2b\)
từ đó bđt cần cm <=> \(a+b+c\ge ab+bc+ca\)
lại có \(ab+bc+ca+abc\le4\)
\(\Leftrightarrow\left(a+2\right)\left(b+2\right)\left(c+2\right)\le\left(a+2\right)\left(b+2\right)+...\)
\(\Leftrightarrow\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\ge1\)
\(\Leftrightarrow\frac{a}{a+2}+\frac{b}{b+2}+\frac{c}{c+2}\le1\)
\(\Leftrightarrow\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2\left(a+b+c\right)}\le\frac{a}{a+2}+\frac{b}{b+2}+\frac{c}{c+2}\le1\)
\(\Rightarrow a+b+c\ge ab+bc+ca\)
=>Q.E.D