CatalinBordea wrote: Let $ x,y,z $ be three positive real numbers such that $ x^2+y^2+z^2+3=2(xy+yz+zx) . $ Show that $$ \sqrt{xy}+\sqrt{yz}+\sqrt{zx}\ge 3, $$and determine in which circumstances equality happens. Vlad Robu We have : $xy+yz+zx=\frac{(x+y+z)^2+3}{4}$ $$LHS=\sum \sqrt{xy}\ge \sum \frac{2xy}{x+y}=2(xy+yz+zx)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x} \right)-2(x+y+z)$$$$LHS \ge \frac{2(xy+yz+zx)}{x+y+z}\left(3+\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \right)-2(x+y+z)$$$$LHS \ge \frac{2(xy+yz+zx)}{x+y+z}\left(3+\frac{(x+y+z)^2}{2(xy+yz+zx)} \right)-2(x+y+z)$$$$LHS \ge \frac{6(xy+yz+zx)}{x+y+z}-(x+y+z)=3+\frac{(x+y+z-3)^2}{2(x+y+z)} \ge 3=RHS$$

With the same conditions $abc\ge 1$.

mudok wrote: With the same conditions $xyz\ge 1$. By Schur's Inequality, $$x^2+y^2+z^2+\frac{9xyz}{x+y+z}\ge 2(xy+yz+zx)$$we deduce $$3xyz\ge x+y+z\ge 3\sqrt[3]{xyz}\Longrightarrow xyz\ge 1\Longrightarrow \sqrt{xy}+\sqrt{yz}+\sqrt{zx}\ge3\sqrt[3]{xyz}\ge3.$$