Solving the system $\Big\{\!\small\begin{array}{l}x+y+z =1 \\ \sqrt{4x+1}+\sqrt{4y+1}+\sqrt{4z+1}=\sqrt{21} \end{array}$

Question

I would be grateful for help in solving the problem.

Question. Solve the system: $$ \begin{cases}x+y+z =1 \\[0.5em] \sqrt{4x+1}+\sqrt{4y+1}+\sqrt{4z+1}=\sqrt{21} \end{cases} $$ with $(x,y,z) \in \mathbb{R}^3$.

I don't understand exactly how to solve this problem, all I know is that it needs to be solved through vectors of three-dimensional space, I tried to set vectors $p(2\sqrt{x},1)$, $ q(2\sqrt{y},1)$, $ l(2\sqrt{z},1)$ and use inequality $$ \sqrt{4x+1}+\sqrt{4y+1}+\sqrt{4z+1}\ge \sqrt{(2\sqrt{x}+2\sqrt{y}+2\sqrt{z})^2+(1+1+1)^2}, $$ but I could not advance further in this way.

Answer 1

The Cauchy-Shwarz inequality, which states that $(\Sigma_{k=1}^n a_kb_k)^2 \le (\Sigma_{k=1}^n a_k^2)(\Sigma_{k=1}^n b_k^2)$, gives for $A = (\sqrt{4x + 1}, \sqrt{4y + 1}, \sqrt{4z + 1})$ and $B = (1, 1, 1)$:

$ \begin{equation} (\sqrt{4x+1} + \sqrt{4y+1} + \sqrt{4z+1})^2 \le (4x + 1 + 4y + 1 + 4z + 1)(3) \\ \iff 21 \le 12x + 12y + 12z + 9 \iff x + y + z \ge 1 \end{equation} $

But since we are given $(1): x + y +z = 1$, we have equality in Cauchy-Schwarz, which occurs if and only if $a_1b_2 = a_2b_1, a_1b_3 = a_3b_1, a_2b_3 = a_3b_2$ (for $n=3$), which gives: $\sqrt{4x + 1} = \sqrt{4y + 1} = \sqrt{4z + 1} \iff x = y = z \implies x = y = z = \frac{1}{3}$.

Answer 2

$x+y+z = 1$

$4x+1 + 4y +1 +4z +1 = 7$

Let $4x + 1 = a^2 , \ 4y + 1 = b^2 ,\ 4z +1 =c^2$

$\therefore \ \ \ a^2 + b^2 + c^2 = 7$

$\implies a + b + c = \sqrt{21}$

$(a + b + c)^2 = 21$

$a^2 + b^2 + c^2 + 2(ab+bc+ca) = 21$

$ab+bc+ca =7$

We know $a^2 + b^2 + c^2 \ge ab + bc+ca$ and equality is held when $a = b= c$

$\therefore a^2 = \frac{7}{3}$

$\implies x = y= z= \frac{1}{3}$

Answer 3

The RMS-AM inequality (root mean square - arithmetic mean) holds for arbitrary real numbers, not necessarily positive, and states that:

$$ \sqrt{\frac{x_1^2+x_2^2+ \dots + x_n^2}{n}} \ge \frac{x_1 + x_2 + \dots x_n}{n} \;\;\;\; \text{with equality iff all } x_k \text{ are equal} $$

Using the inequality for $n=3$ and $x_1=\sqrt{4x+1}$, $x_2=\sqrt{4y+1}$, $x_3=\sqrt{4z+1}\,$:

$$ \sqrt{\frac{7}{3}} = \sqrt{\frac{4 \cdot \color{red}{1} + 3}{3}} = \sqrt{\frac{4 (\color{red}{x+y+z})+ 3}{3}} \ge \frac{\color{blue}{\sqrt{4x+1}+\sqrt{4y+1}+\sqrt{4z+1}}}{3} = \frac{\color{blue}{\sqrt{21}}}{3} $$

Since $\sqrt{\frac{7}{3}} = \frac{\sqrt{21}}{3}$ the two sides of the inequality are equal, and it follows that this must be the equality case of the RMS-AM inequality, so $x=y=z$.