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For all real numbers $a$ and $b$ find the minimum of the following expression.
$$(a-b)^2 + (2-a-b)^2 + (2a-3b)^2$$

I tried expressing the entire expression in terms of a single function of $a$ and $b$. For example, if the entire expression reduces to $(a-2b)^2+(a-2b)+5$ then its minimum can be easily found. But nothing seems to get this expression in such a form, because of the third unsymmetric square.

Since there are two variables here we can also not use differentiation.

Can you please provide hints on how to solve this?

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  • $\begingroup$ Hint : First check whether all the paranthesis can be $0$. If yes, the global minimum must be $0$. Unfortunately, this is not the case here. $\endgroup$
    – Peter
    Commented Jul 28, 2017 at 13:37
  • $\begingroup$ what kind of numbers are $a,b$? $\endgroup$
    – Dr. Sonnhard Graubner
    Commented Jul 28, 2017 at 13:39
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    $\begingroup$ You can use differentiation. Try using partial derivative. $\endgroup$
    – MCCCS
    Commented Jul 28, 2017 at 13:43
  • $\begingroup$ @MCCCS Can we solve it without using partial derivatives? We haven't been taught that. $\endgroup$
    – Arishta
    Commented Jul 28, 2017 at 13:47
  • $\begingroup$ wolframalpha.com/input/… $\endgroup$
    – Peter
    Commented Jul 28, 2017 at 14:02

3 Answers 3

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\begin{eqnarray*} (a-b)^2+(2-a-b)^2+(2a-3b)^2=6a^2-12ab+11b^2-4(a+b)+4 \\ =6\left(a-b-\frac{1}{3}\right)^2+5\left(b-\frac{4}{5}\right)^2+\color{red}{\frac{2}{15}}. \end{eqnarray*}

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  • $\begingroup$ How did you get the idea that the expression can be written in form of these two squares. I tried a lot of arrangements but couldn't form these two squares. $\endgroup$
    – Arishta
    Commented Jul 28, 2017 at 13:51
  • $\begingroup$ @Donald Splutterwit There is a mistake in your solution. See my proof. It's right! $\endgroup$
    – Michael Rozenberg
    Commented Jul 28, 2017 at 13:51
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    $\begingroup$ @Eloise The first term is obtained from completing the square for $a$, so that you're only left with constant terms and terms in powers of $b$. Then you complete the square for $b$, and this leaves you with a constant term, which must be the minimum value. By setting each of the squared terms to zero, you can then find $a$ and $b$. $\endgroup$
    – Tez LaCoyle
    Commented Jul 28, 2017 at 13:52
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    $\begingroup$ @Donald Splutterwit The mistake Michael Rozenberg refers to is in the LHS of the first line - you've mis-transcribed the last term, which was $(2a-3b)^{2}$ as opposed to $(2a+3b)^{2}$. The expansion and subsequent steps are correct though. $\endgroup$
    – Tez LaCoyle
    Commented Jul 28, 2017 at 13:54
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    $\begingroup$ @MichaelRozenberg ... well spotted ... edit now ...Thanks. $\endgroup$
    – Donald Splutterwit
    Commented Jul 28, 2017 at 13:57
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Let $a=\frac{17}{15}$ and $b=\frac{4}{5}$.

Hence, we get a value $\frac{2}{15}$.

Thus, it remains to prove that $$(a-b)^2 + (2-a-b)^2 + (2a-3b)^2\geq\frac{2}{15}$$ or $$10(3a-3b-1)^2+3(5b-4)^2\geq0$$ Done!

I got my solution by the following way.

We need to find a maximal $k$ for which the following inequality is true for all reals $a$, $b$ and $c$. $$(a-b)^2 + (2-a-b)^2 + (2a-3b)^2\geq k$$ or $$6a^2-4(3b+1)a+11b^2-4b+4-k\geq0,$$ for which we need $$4(3b+1)^2-6(11b^2-4b+4-k)\leq0$$ or $$15b^2-24b+10-3k\geq0,$$ for which we need $$12^2-15(10-3k)\leq0$$ or $$k\leq\frac{2}{15}.$$ The equality occurs for $k=\frac{2}{15}$, $b=\frac{24}{2\cdot15}$, which is $b=\frac{4}{5}$ and for these values we obtain $$(a-b)^2 + (2-a-b)^2 + (2a-3b)^2\geq \frac{2}{15}$$ it's $$6a^2-4(3b+1)a+11b^2-4b+4-\frac{2}{15}\geq0$$ or $$90a^2-60(3b+1)a+165b^2-60b+58\geq0$$ or $$10(9a^2-6(3b+1)a+(3b+1)^2)-10(3b+1)^2+165b^2-60b+58\geq0$$ or $$10(3a-3b-1)^2+75b^2-120b+48\geq0$$ or $$10(3a-3b-1)^2+3(5a-4)^2\geq0.$$

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    $\begingroup$ You're starting with the answer though. That doesn't really help OP to solve the problem, all it does is verify that $a=\frac{17}{15}, b=\frac{4}{5}$ is the solution. $\endgroup$
    – Tez LaCoyle
    Commented Jul 28, 2017 at 13:50
  • $\begingroup$ @T. Linnell See better my solution, please. My solution it's exactly like Donald Splutterwit wrote, but I made it before and without mistakes. $\endgroup$
    – Michael Rozenberg
    Commented Jul 28, 2017 at 13:55
  • $\begingroup$ You're starting from $a=\frac{17}{15}, b=\frac{4}{5}$ with no explanation of how you've chosen these values. Your solution is correct, and you did post it first, but as written, it only solves the problem, without really explaining how to. $\endgroup$
    – Tez LaCoyle
    Commented Jul 28, 2017 at 14:00
  • $\begingroup$ @T. Linnell I just proved the quadratic inequality. Should be $\Delta\leq0$ and from here we can get the equality case $a=\frac{17}{5}$ and $b=\frac{4}{5}$. I think Donald Splutterwit used the same way. Otherwise, it's impossible to get his and my solution. $\endgroup$
    – Michael Rozenberg
    Commented Jul 28, 2017 at 14:04
  • $\begingroup$ @Eloise I posted a full solution for you. $\endgroup$
    – Michael Rozenberg
    Commented Jul 28, 2017 at 14:33
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$$(a-b)^2 + (2-a-b)^2 + (2a-3b)^2 = \left\| \,\, \begin{bmatrix} 1 & -1\\ 1 & 1\\ 2 & -3\end{bmatrix} \begin{bmatrix} a\\ b\end{bmatrix} - \begin{bmatrix} 0\\ 2\\ 0\end{bmatrix} \,\, \right\|_2^2$$

This is a least-squares problem. Since the matrix has full column rank, the minimum is

$$\left\| \,\, \begin{bmatrix} 1 & -1\\ 1 & 1\\ 2 & -3\end{bmatrix} \left( \begin{bmatrix} 1 & -1\\ 1 & 1\\ 2 & -3\end{bmatrix}^\top \begin{bmatrix} 1 & -1\\ 1 & 1\\ 2 & -3\end{bmatrix} \right)^{-1} \begin{bmatrix} 1 & -1\\ 1 & 1\\ 2 & -3\end{bmatrix}^\top \begin{bmatrix} 0\\ 2\\ 0\end{bmatrix} - \begin{bmatrix} 0\\ 2\\ 0\end{bmatrix} \,\, \right\|_2^2 = \color{blue}{\frac{2}{15}}$$

SymPy code

>>> from sympy import *
>>> A = Matrix([[ 1,-1],
                [ 1, 1],
                [ 2,-3]])
>>> b = Matrix([0,2,0])
>>> error = A * (A.T * A)**-1 * A.T * b - b

The squared Euclidean norm of the error vector is

>>> error.T * error
Matrix([[2/15]])
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