XIV APMO

$$\begin{array}{rl}\sum _{\text{cyclic}}\sqrt{x+yz}& =\sqrt{xyz}\sum _{\text{cyclic}}\sqrt{\frac{1}{x}+\frac{1}{yz}}\\ & =\sqrt{xyz}\sum _{\text{cyclic}}\sqrt{\frac{1}{x}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+\frac{1}{yz}}[1\text{mark.}]\\ & =\sqrt{xyz}\sum _{\text{cyclic}}\sqrt{\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x}+\frac{1}{z}\right)}[1\text{mark.}]\end{array}$$ $$\begin{array}{rl}& =\sqrt{xyz}\sum _{\text{cyclic}}\sqrt{{\left(\frac{1}{x}÷\frac{1}{\sqrt{yz}}\right)}^2+\frac{(\sqrt{y}-\sqrt{z}{)}^2}{xyz}}[2\text{marks.}]\\ & \ge \sqrt{xyz}\sum _{\text{cyclic}}\left(\frac{1}{x}+\frac{1}{\sqrt{yz}}\right)[1\text{mark.}]\\ & =\sqrt{xyz}\left(1+\sum _{\text{cyclic}}\frac{1}{\sqrt{yz}}\right)[1\text{mark.}]\\ & =\sqrt{xyz}+\sum _{\text{cyclic}}\sqrt{x}.[1\text{mark.}]\end{array}$$

Squaring both sides of the given inequality, we obtain $$\begin{array}{rl}& \sum _{\text{cyclic}}x+\sum _{\text{cyclic}}yz+2\sum _{\text{cyclic}}\sqrt{x+yz}\sqrt{y+zx}\\ & \ge xyz+2\sqrt{xyz}\sum _{\text{cyclic}}\sqrt{x}+\sum _{\text{cyclic}}x+2\sum _{\text{cyclic}}\sqrt{xy}.[1\text{mark.}]\end{array}$$ It follows from the given condition $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ that $xyz={\sum }_{\text{cyclic}}xy$. Therefore, the given inequality is equivalent to $$\sum _{\text{cyclic}}\sqrt{x+yz}\sqrt{y+zx}\ge \sqrt{xyz}\sum _{\text{cyclic}}\sqrt{x}+\sum _{\text{cyclic}}\sqrt{xy}.\text{[2 marks.]}$$ Using the Cauchy-Schwarz inequality [or just $x^2+y^2\ge 2xy$], we see that $$(x+yz)(y+zx)\ge {\left(\sqrt{xy}+\sqrt{xy{z}^2}\right)}^2,[1\text{mark.}]$$ or $$\sqrt{x+yz}\sqrt{y+zx}\ge \sqrt{xy}+\sqrt{z}\sqrt{xyz}.[1\text{mark.}]$$ Taking the cyclic sum of this inequality over $x$, $y$ and $z$, we get the desired inequality. [2 marks.]

Using the condition $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ and the AM-GM inequality, we have $$\begin{array}{rl}x+yz-{\left(\sqrt{\frac{yz}{x}}+\sqrt{x}\right)}^2& =yz\left(1-\frac{1}{x}\right)-2\sqrt{yz}\\ & =yz\left(\frac{1}{y}+\frac{1}{z}\right)-2\sqrt{yz}=y+z-2\sqrt{yz}\ge 0\end{array}$$ which gives $$\sqrt{x+yz}\ge \sqrt{\frac{yz}{x}}+\sqrt{x}.[3\text{marks.}]$$ Similarly, we have $$\sqrt{y+zx}\ge \sqrt{\frac{zx}{y}}+\sqrt{y}\text{and}\sqrt{z+xy}\ge \sqrt{\frac{xy}{z}}+\sqrt{z}.$$ Addition yields $$\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge \sqrt{\frac{yz}{x}}+\sqrt{\frac{zx}{y}}+\sqrt{\frac{xy}{z}}+\sqrt{x}+\sqrt{y}+\sqrt{z}.$$ [2 marks.] Using the condition $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ again, we have $$\sqrt{\frac{yz}{x}}+\sqrt{\frac{zx}{y}}+\sqrt{\frac{xy}{z}}=\sqrt{xyz}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\sqrt{xyz},[1\text{mark.}]$$ and thus $$\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge \sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}.[1\text{mark.}]$$

We make the substitution $a=\frac{1}{x}$, $b=\frac{1}{y}$, $c=\frac{1}{z}$. Then it is enough to show that $$\sqrt{\frac{1}{a}+\frac{1}{bc}}+\sqrt{\frac{1}{b}+\frac{1}{ca}}+\sqrt{\frac{1}{c}+\frac{1}{ab}}\ge \sqrt{\frac{1}{abc}}+\sqrt{\frac{1}{a}}+\sqrt{\frac{1}{b}}+\sqrt{\frac{1}{c}}$$ where $a+b+c=1$. Multiplying this inequality by $\sqrt{abc}$, we find that it can be written $$\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}\ge 1+\sqrt{bc}+\sqrt{ca}+\sqrt{ab}.[1\text{mark.}]$$ This is equivalent to $$\begin{array}{rl}\sqrt{a(a+b+c)+bc}+\sqrt{b(a+b+c)+ca}+\sqrt{c(a+b+c)+ab}& \\ \ge a+b+c+\sqrt{bc}+\sqrt{ca}+\sqrt{ab},& \{[1\text{mark.}]\}\end{array}$$ which in turn is equivalent to $$\sqrt{(a+b)(a+c)}+\sqrt{(b+c)(b+a)}+\sqrt{(c+a)(c+b)}\ge a+b+c+\sqrt{bc}+\sqrt{ca}+\sqrt{ab}$$ [1 mark.] (This is a homogeneous version of the original inequality.) By the Cauchy-Schwarz inequality (or since $b+c\ge 2\sqrt{bc}$), we have $$\left[(\sqrt{a}{)}^2+(\sqrt{b}{)}^2\right]\left[(\sqrt{a}{)}^2+(\sqrt{c}{)}^2\right]\ge (\sqrt{a}\sqrt{a}+\sqrt{b}\sqrt{c}{)}^2$$ or $$\sqrt{(a+b)(a+c)}\ge a+\sqrt{bc}.[2\text{marks.}]$$ Taking the cyclic sum of this inequality over $a$, $b$, $c$, we get the desired inequality. [2 marks.]