To prove this inequality, we can use the AM-GM inequality, which states that for any positive real numbers $a$ and $b$, $\frac{a+b}{2}\ge\sqrt{ab}$. We can apply this inequality to the three terms $xy$, $yz$, and $zx$ to get $\frac{xy+yz+zx}{3}\ge\sqrt[3]{(xy)(yz)(zx)}=xyz$. Multiplying both sides by $2$ gives us $xy+yz+zx\ge2xyz$. Combining this with the fact that $x+y+z=1$, we get $xy+yz+zx-2xyz\ge0$. To prove the upper bound of $\frac{7}{27}$, we can use the fact that $x+y+z=1$ to rewrite the inequality as $xy+yz+zx\le\frac{1}{3}-2xyz$. Then, using the AM-GM inequality again, we get $\frac{xy+yz+zx}{3}\le\frac{(x+y+z)^2}{9}=\frac{1}{9}$. Combining this with the fact that $x+y+z=1$, we get $xy+yz+zx\le\frac{1}{3}-2xyz\le\frac{1}{3}-2\left(\frac{1}{9}\right)=\frac{7}{27}$. Therefore, we have proven the inequality $0\le xy+yz+zx-2xyz\le\frac{7}{27}$ for $x+y+z=1$.
The AM-GM inequality is a fundamental inequality in mathematics that states that for any positive real numbers $a$ and $b$, the arithmetic mean (AM) of $a$ and $b$ is always greater than or equal to the geometric mean (GM) of $a$ and $b$. In other words, $\frac{a+b}{2}\ge\sqrt{ab}$. This inequality can be used to prove the given inequality by applying it to the three terms $xy$, $yz$, and $zx$. By multiplying both sides of the inequality by $2$, we can rearrange the terms to get $xy+yz+zx\ge2xyz$. This is the lower bound of the given inequality. To prove the upper bound, we can rewrite the inequality as $xy+yz+zx\le\frac{1}{3}-2xyz$ and then apply the AM-GM inequality again to get $\frac{xy+yz+zx}{3}\le\frac{(x+y+z)^2}{9}=\frac{1}{9}$. Combining this with the fact that $x+y+z=1$, we get $xy+yz+zx\le\frac{1}{3}-2xyz\le\frac{1}{3}-2\left(\frac{1}{9}\right)=\frac{7}{27}$. Therefore, the AM-GM inequality is a useful tool in proving the given inequality.
Yes, the given inequality is always true for any values of $x$, $y$, and $z$ that satisfy $x+y+z=1$. This can be proven using the AM-GM inequality, as explained in the first question. Since the AM-GM inequality is a fundamental principle in mathematics, it holds true for all positive real numbers, which includes all possible values of $x$, $y$, and $z$ that satisfy $x+y+z=1$.
One example of values for $x$, $y$, and $z$ that satisfy $x+y+z=1$ and also satisfy the given inequality is $x=\frac{1}{3}$, $y=\frac{1}{3}$, and $z=\frac{1}{3}$. Plugging these values into the given inequality, we get $\frac{1}{9}+\frac{1}{9}+\frac{1}{9}-2\left(\frac{1}{27}\right)=\frac{1