Lagrange's revenge

After some simplification, the given expression becomes (a-c) ^2+(a+3ln(a)-2c-3)^2. Differentiating partially with respect to a and c, and equating to zero we get a=3. The value of the expression then becomes (c-3)^2+(3ln3-2c)^2. This attains a minimum value of 1.8(ln3-2)^2=1.462499651052258

Let $f(a,b,c,d)=(a-c)^2+(b-d)^2, \quad g_1(a,b,c,d)=a+3\ln a - b, \quad g_2(a,b,c,d)=d-2c.$

Using Lagrange multipliers,
$\begin{aligned} \nabla f &=\lambda_1\nabla g_1+\lambda_2\nabla g_2 \\ (2(a-c),2(b-d),-2(a-c),-2(b-d))&=\lambda_1(1+\frac3a,-1,0,0)+\lambda_2(0,0,-2,1) \\ \frac{2(a-c)}{1+\frac3a}&=-2(b-d) \\ a-c&=-2(b-d) \\ \end{aligned}$
We also have $d-3=2c$ and $a+3\ln a=b$ .

If the stationary point found was a saddle point, then min $(f)=-\infty$ which is impossible. The stationary point also can't be a maximum, since $a$ and $c$ don't depend on each other so $(a-c)^2$ can be as large as we want. Thus, the stationary point must be a minimum, and solving the last $4$ equations for $a,b,c,d$ we find min $(f)=\frac95(2-\ln3)^2\approx1.462499651052258$ .