finding $(x,y,z,t)$ in system of equation

Question

If $x,y,z,t$ are positive real number and $x+y+z+t=12$ and $xyzt=27+xy+xz+xt+yz+yt+zt$.then $(x,y,z,t)$ are

solution i try arithmetic geometric inequality $xyzt=27+6\sqrt[2]{xyzt}$

Help me to solve from that point

Answer 1

You're right—the arithmetic–geometric mean inequality is the way to go. Let $A$ be the arithmetic mean of $x$, $y$, $z$, and $t$, and let $G$ be their geometric mean. Then the first equation is $$A=3$$ which by the arithmetic–geometric mean inequality gives $$G\leq 3$$ and the second gives the inequality (not an equality) $$G^4\geq 27+6G^2\text{.}$$ This inequality—along with $G\geq 0$—implies that $G\geq 3$.

So we know that $A=G=3$. But the converse of the arithmetic–geometric mean inequality states that these two means are equal if and only if all of their terms are equal. So we know that $$\boxed{x=y=z=t=3}.$$