IRL_ABooklet_2023

Solution 1. Adding $3$ to each side, we have to show $$\begin{array}{rl}5& \le \frac{a+b}{c+2}+1+\frac{b+c}{a+2}+1+\frac{c+a}{b+2}+1\\ & =(a+b+c+2)\left(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\right)\\ & =5\left(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\right)\end{array}$$ Dividing by $5$, this is equivalent to $$\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\ge 1(4)$$ Alternatively, we may substitute $a+b=3-c=5-(c+2)$ in the first numerator and so on, which again leads us to (4). Now Jensen's inequality applied to the convex function $f(x)=\frac{1}{x+2}$ at the points $a$, $b$, $c$ implies: $$\frac{1}{3}\left(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\right)\ge \frac{1}{\frac{1}{3}(a+b+c)+2}=\frac{1}{3}.$$ Equality holds (in general) when $a=b=c$ and in this specific case when $a=b=c=1$ since we are given $a+b+c=3$.

Rather than invoking Jensen's inequality, (4) can be proved by appeal to the AM-HM inequality, the AM-GM inequality or the Cauchy-Schwarz inequality. The AM-HM inequality application is immediate after multiplication by $(a+2)+(b+2)+(c+2)=9$. The use of AM-GM requires two steps: $$\begin{array}{rl}\frac{1}{3}\left(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\right)& \ge \frac{1}{\sqrt[3]{(a+2)(b+2)(c+2)}}\\ & \ge \frac{3}{a+2+b+2+c+2}=\frac{1}{3}.\end{array}$$ The Cauchy-Schwarz approach applies the inequality to the vectors $$\left(\sqrt{a+2},\sqrt{b+2},\sqrt{c+2}\right)\text{and}\left(\frac{1}{\sqrt{a+2}},\frac{1}{\sqrt{b+2}},\frac{1}{\sqrt{c+2}}\right)$$ whence $$\begin{array}{rl}9& \le ((a+2)+(b+2)+(c+2))\left(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\right)\\ & =9\left(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\right).\end{array}$$

$$\frac{a+b}{5-a-b}+\frac{3-a}{a+2}+\frac{3-b}{b+2}\ge 2.$$ Cross-multiplying and moving all terms to the left (noting all the denominators are positive) gives: $5a^{2}b+5a{b}^2+5a^2+5b^2-10ab-15a-15b+20\ge 0$. The left hand side can be rearranged as $5b(a-1{)}^2+5a(b-1{)}^2+5(a+b-2{)}^2$. This is non-negative, and zero iff $a=b=1$.