If $x,y,z,t\geq 0,$ Then $\max(xy+yz+zt)$

Question

If $x,y,z,t\geq 0$ and $x+y+z+t=1\;,$ Then $\max$ value of $(xy+yz+zt)$

$\bf{My\; Try::}$ We can write $x+y+z+t=(x+z)+(y+t)\geq 2(x+z)(y+t)$

and equality hold when $x+z=y+t$. So we get $\displaystyle x+z=y+t=\frac{1}{2}$

So we get $$\displaystyle (x+z)(y+t)\leq \frac{1}{4}\Rightarrow xy+yz+zt+xt\leq \frac{1}{4}$$

Now how can i solve it after that, help required, Thanks

Answer 1

Because $(x+y+z+t)^2-4(xy+yz+zt)=(x-y+z-t)^2+4xt\geq0$

and the equality occurs for example, for $x=y$ and $t=z=0$.