Inequality in Triangle: Prove $|x^2+y^2+z^2-2(xy+yz+xz)|\le \frac{1}{27}$

[maxkor]

Let $x, y, z$ be length of the side of a triangle such that $\sqrt{x} + \sqrt{y} + \sqrt{z} = 1.$
Prove $|x^{2} + y^{2} + z^{2} - 2\left( xy+yz+xz\right)| \le \frac{1}{27}$.

 

[jvicens]

I would like to provide a proof for the inequality $|x^{2} + y^{2} + z^{2} - 2\left( xy+yz+xz\right)| \le \frac{1}{27}$, given that $\sqrt{x} + \sqrt{y} + \sqrt{z} = 1$.

First, we can rewrite the given condition as $\sqrt{x} + \sqrt{y} + \sqrt{z} = \sqrt[3]{1}$. This suggests that we can use the AM-GM inequality, which states that for positive real numbers $a_1, a_2, ..., a_n$, $\frac{a_1 + a_2 + ... + a_n}{n} \ge \sqrt[n]{a_1a_2...a_n}$.

Applying this to our three variables, we have $\frac{\sqrt{x} + \sqrt{y} + \sqrt{z}}{3} \ge \sqrt[3]{\sqrt{x}\sqrt{y}\sqrt{z}}$. Simplifying, we get $\frac{1}{3} \ge \sqrt[6]{xyz}$. Squaring both sides, we have $\frac{1}{9} \ge xyz$.

Next, we can use the Cauchy-Schwarz inequality, which states that for any $n$ real numbers $a_1, a_2, ..., a_n$ and $n$ real numbers $b_1, b_2, ..., b_n$, $(a_1b_1 + a_2b_2 + ... + a_nb_n)^2 \le (a_1^2 + a_2^2 + ... + a_n^2)(b_1^2 + b_2^2 + ... + b_n^2)$.

Applying this to our three variables, we have $(x+y+z)^2 \le (x^2 + y^2 + z^2)(1+1+1)$. Simplifying, we get $x^2 + y^2 + z^2 \ge \frac{(x+y+z)^2}{3}$.

Using the given condition, we can substitute $x + y + z = 1 - \sqrt{x} - \sqrt{y} - \sqrt{