Solve the system of equations: $ \sqrt{3(x - y)^2 - 4x + 8y + 5} - \sqrt x = \sqrt{y + 1}$ and $x^2y + y^2 - 3xy - 3x + 7y + 8 = 2x\sqrt{y + 3}$.

Question

Solve the system of equations $$\large \left\{ \begin{align} \sqrt{3(x - y)^2 - 4x + 8y + 5} - \sqrt x = \sqrt{y + 1}\\ x^2y + y^2 - 3xy - 3x + 7y + 8 = 2x\sqrt{y + 3} \end{align} \right.$$

This is a problem that heavily uses inequation to solve. I have provided my answer below, which looks like a mess of its own.

Answer 1

Let $y' = y + 1$ $(x, y' \ge 0)$, the system of equations become

$$\left\{ \begin{align} \sqrt{3(x - y')^2 + 2x + 2y'} - \sqrt x = \sqrt{y'}\\ x^2y' - x^2 + y'^2 - 3xy' + 5y' + 2 = 2x\sqrt{y' + 2} \end{align} \right.$$

We have that $\sqrt{3(x - y')^2 + 2x + 2y'} \ge \sqrt{2\left[(\sqrt x)^2 + (\sqrt{y'})^2\right]} \ge \sqrt x + \sqrt{y'}$

The inequality occurs when $x = y'$ $(\iff x = y - 1)$.

Plugging in $x = y'$ into the second equation, we have that $x^3 - 3x^2 + 5x + 2 = 2x\sqrt{x + 2}$

$x^3 - 3x^2 + 5x + 2 = 2x\sqrt{x + 2} = x(x - 2)^2 + x^2 + (x + 2) \ge 2\sqrt{x^2(x + 2)} = 2x\sqrt{x + 2}$

The equality sign occurs when $x = 2 \implies y = 1$.

Answer 2

You can write your idea without substitution.

By C-S we obtain: $$\sqrt{x}+\sqrt{y+1}=\sqrt{3(x-y)^2-4x+8y+5}=\sqrt{3(x-y-1)^2+6(x-y)-3-4x+8y-5}=$$ $$=\sqrt{3(x-y-1)^2+2x+2(y+1)}\geq\sqrt{(1^2+1^2)(x+y+1)}\geq\sqrt{x}+\sqrt{y+1}.$$ The equality occurs for $x-y=1=0$ and $(1,1)||(\sqrt{x},\sqrt{y+1})$, which gives $$y=x-1$$ and the rest is smooth.