Irish Mathematical Olympiad

Solution to Part (a). One can establish this in several ways. For instance, using the convexity of $t\to t^2$ we have $${\left(\frac{3+2(x+y+z)}{3}\right)}^2={\left(\frac{(1+x+y)+(1+y+z)+(1+z+x)}{3}\right)}^2\le \frac{(1+x+y{)}^2+(1+y+z{)}^2+(1+z+x{)}^2}{3},$$ with equality iff $$1+x+y=1+y+z=1+z+x,$$ i.e., $x=y=z$. Hence $$\frac{(3+2(x+y+z){)}^2}{3}\le (1+x+y{)}^2+(1+y+z{)}^2+(1+z+x{)}^2,$$ with equality iff $x=y=z$. This also follows from an application of the Cauchy-Schwarz inequality to the expression $$3+2(x+y+z)=(1+x+y)+(1+y+z)+(1+z+x).$$ To continue, by the AM-GM inequality, $$3+2(x+y+z)\ge 3+6\sqrt{xyz}\ge 9,$$ with equality iff $x=y=z=1$. Hence $$27=\frac{81}{3}\le \frac{(3+2(x+y+z){)}^2}{3}\le (1+x+y{)}^2+(1+y+z{)}^2+(1+z+x{)}^2,$$ with equality iff $x=y=z$. Alternatively, applying the AM-GM inequality termwise we see that $$\begin{array}{rl}(1+x+y{)}^2+(1+y+z{)}^2+(1+z+x{)}^2& \ge (3\sqrt[3]{1xy}{)}^2+(3\sqrt[3]{1yz}{)}^2+(3\sqrt[3]{1zx}{)}^2\\ & =9(\sqrt[3]{x^2{y}^2}+\sqrt[3]{y^2{z}^2}+\sqrt[3]{z^2{x}^2})\\ & \ge 27\sqrt[3]{x^4{y}^4{z}^4}\\ & \ge 27,\end{array}$$ with equality throughout iff $x=y=z=1$.

Solution to Part (b). Now consider the inequality on the right-hand side: $$(1+x+y{)}^2+(1+y+z{)}^2+(1+z+x{)}^2\le 3(x+y+z{)}^2.$$ Expanding and simplifying this leads to the following equivalent statement that $$3+2(x^2+y^2+z^2)+2(xy+yz+zx)+4(x+y+z)\le 3(x^2+y^2+z^2)+6(xy+yz+zx),$$ i.e., $$3+4(x+y+z)\le x^2+y^2+z^2+4(xy+yz+zx)=(x+y+z{)}^2+2(xy+yz+zx).$$ Equivalently, $$7\le (u-2{)}^2+2v,$$ where $u=x+y+z$, $v=xy+yz+zx$. But $u\ge 3$ and $v\ge 3$, with equality in both of these inequalities iff $x=y=z=1$. Hence $$(u-2{)}^2+2v\ge 1+6=7,$$ with equality iff $x=y=z=1$. This establishes the desired result.