$\sqrt{y-1} \leq \frac{y-1 + 1}{2} = \frac{y}{2}$

Equality only occurs when $x=y=z=2$. Just wanted to point that out.

Using what mahanamth said we conclude that the only possible solution is equality and that is x=y=z=2

WakeUp wrote: Find all the triples $(x,y,z)$ of real numbers such that \[2x\sqrt{y-1}+2y\sqrt{z-1}+2z\sqrt{x-1} \ge xy+yz+zx \] We have $x,y,z \ge\ 1$.Now lets $y = a^2 + 1,x = b^2 + 1,z = c^2 + 1$.If we replace a,b,c insteed of x,y,z we obtain $2a(b^2+1) + 2c(a^2+1) + 2b(c^2+1) \ge\ a^2b^2 + b^2c^2 + c^2a ^ 2 + 2a^2 + 2b^2 + 2c^2 + 3$,but we have $a^2 + 1 \ge\ 2a$ and $b^2 + a^2b^2 \ge\ 2ab^2$ $\implies a^2 + 1 = 2a$ $\implies a = b = c = 1$.We get that $x = y = z = 2$.Nice problem!!

Nice? there are oodles of such questions, based on the equality case of some inequality. More like ... pathetic

mahanmath wrote: $\sqrt{y-1} \leq \frac{y-1 + 1}{2} = \frac{y}{2}$ Continuing, then we can substitute this into the original to get $2xy/2+2yz/2+2zx/2\geq$ \[2x\sqrt{y-1}+2y\sqrt{z-1}+2z\sqrt{x-1} \ge xy+yz+zx \], meaning the equality case must occur, or only x=y=z=2.