Find $a$ for which $y\ge x^2+4a^2,x\ge y^2+4a^2$ has only one solution For what $a$ does this system of inequalities have only one solution?
$$
\begin{cases}y\ge x^2+4a^2\\x\ge y^2+4a^2\end{cases}
$$
So, I have tried getting $y$ from the first equation then putting it into second, but then I have two quadratic and $x$, which of these quadratics are the first term? I know that I have to put it into $\Delta=0$. But don't really get which of them are first, second etc.
Even tried subtracting second to first.
 A: If $(x,y)$ is a solution so is $(y,x)$, so for one solution we need $x=y$ – we thus seek $a$ for which $y=x^2+4a^2$ is tangent to $y=x$, i.e. $x^2-x+4a^2$ has one solution, hence zero discriminant. Here $\Delta=1-16a^2=0$, so $a=\pm\frac14$ where the only solution is $(x,y)=(1/2,1/2)$.
A: If $(x_0,y_0), x_0\ne y_0$ is a solution then $(y_0,x_0)$ is also a solution. So in order to have only one solution, one must have $x=y$. Plugging $y=x$ in the first equation one gets:
$$x^2-x+4a^2\le 0$$
Which has an unique solution iff $\Delta=1-(4a)^2=0$ which is equivalent to $a=\pm\frac{1}{4}$
A: We have,
$$\begin{align}&\begin{cases}y^2≥\left(x^2+4a^2\right)^2≥0\\y^2≤x-4a^2,\thinspace x≥4a^2\end{cases}\\\\
\implies &\left(x^2+4a^2\right)^2≤y^2≤x-4a^2\\
\implies &\left(x^2+4a^2\right)^2≤x-4a^2 \\
\implies &16a^4+a^2(8x^2+4)+(x^4-x)≤0\\
\implies &\left(4a^2+\left(x^2+\frac 12\right)\right)^2≤\left(x+\frac 12\right)^2\\
\implies &x^2-x+4a^2≤0\\
\implies &\left(x-\frac 12\right)^2≤\frac{1-16a^2}{4}\end{align}$$
Similary, we have
$$\begin{align}\left(y^2+4a^2\right)^2≤y-4a^2\\
\implies \left(y-\frac 12\right)^2≤\frac{1-16a^2}{4}\end{align}$$
We see that, if $x,y$ are unique then
$$16a^2=1\iff a=±\frac 14$$
which implies,
$$x=y=\frac 12.$$
