IRL_ABooklet_2023

We may homogenise in two steps the inequality we wish to show. First, we use $(a+b+c{)}^2=9$ to replace the RHS and obtain the equivalent inequality $$a^2+b^2+c^2+6abc\le (a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)+3abc.$$ This is equivalent to $3abc\le ab+bc+ca$. We now multiply by $3=a+b+c$ to obtain the equivalent inequality $$9abc\le (a+b+c)(ab+bc+ca)=a^{2}b+a{b}^2+b^{2}c+b{c}^2+c^{2}a+a^{2}c+3abc,$$ i.e., $$6abc\le a^{2}b+a{b}^2+b^{2}c+b{c}^2+c^{2}a+a^{2}c,\text{ or}$$ $$6\le \frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c},$$ which immediately follows from the AM-GM inequality.