Cauchy Schwarz - Finding minimum of a function

Question

For $x, y , z$ in real numbers, If $2x+y+z=5$, then what is the min value of $x^2 + y^2 + z^2 - 2x + 4y + 6$. This is a weekend brain teaser in the 2nd week of Calc 3.

Answer 1

This problem can be solved by using elementary math only (no derivatives, no Lagrange multipliers, no C-S ineqaulity etc).

We have to minimize the expression:

$$f=x^2+y^2+z^2-2x+4y+6=(x-1)^2+(y+2)^2+z^2+1$$

Introduce substitution:

$$u=x-1, \ \ v=y+2, \ \ w=z$$

The expression that has to be minimized now becomes:

$$f=u^2+v^2+w^2+1\tag{1}$$

...with the following constraint:

$$2u+v+w=5\tag{2}$$

Make one more subsitution:

$$v=x-y, \ \ w=x+y$$

Replace that into (1) and (2) and you get:

$$f=u^2+2x^2+2y^2+1\tag{3}$$

...and the constraint becomes:

$$2u+2x=5\tag{4}$$

The point is: contraint (4) give us the possibility to minimize (3) by simply choosing $y=0$. Now the problem is much simpler. We have to minimize:

$$f=u^2+2x^2+1\tag{5}$$

...with constraint:

$$u+x=\frac52\tag{6}$$

From (6):

$$x=\frac 52-u$$

...so (7) becomes:

$$f=u^2+2(\frac 52 - u)^2+1=3(u-\frac53)^2+\frac{31}{6}$$

So the minimum value is $f_{min}=\frac{31}{6}$ and it is reached for $u=\frac53$.

Answer 2

Here is an alternate solution using the Cauchy-Schwarz Inequality:-

We have to minimize $f(x,y,z)=x^2+y^2+z^2-2x+4y+6=(x-1)^2+(y+2)^2+z^2+1$ subject to $2x+y+z=5$

From the Cauchy-Schwarz Inequality , we have $$(2\cdot(x-1)+1 \cdot (y+2) +1\cdot z)^2 \leq (2^2+1^2+1^2)((x-1)^2+(y+2)^2+z^2)$$ $$ \therefore (2x+y+z)^2 \leq 6\cdot ( (f(x,y,z)-1)$$

$$ \implies \frac{(2x+y+z)^2}{6} + 1 \leq f(x,y,z) \implies f(x,y,z) \geq \frac{25}{6}+1=\boxed{\frac{31}{6}} $$

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Let’s take it a step further , and generalise the result .

We have to find the minimum of $f(x,y,z)=a_1x^2+b_1y^2+c_1z^2+a_2x+b_2y+c_2z+d$ , subject to $a_0x+b_0y+c_0z=l$

First , we complete the squares , and obtain $$f(x,y,z)=(\sqrt{a_1}x+\frac{a_2}{2\sqrt{a_1}})^2+(\sqrt{b_1}y+\frac{b_2}{2\sqrt{b_2}})^2+(\sqrt{c_1}z+\frac{c_2}{2\sqrt{c_1}})^2 +(d-(\frac{a_2^2}{4a_1}+\frac{b_2^2}{4b_1}+\frac{c_2^2}{4c_1}))$$

Then , by the Cauchy Schwarz Inequality , we have :- $$((\sqrt{a_1}x+\frac{a_2}{2\sqrt{a_1}})\cdot \frac{a_0}{\sqrt{a_1}}+(\sqrt{b_1}y+\frac{b_2}{2\sqrt{b_1}})\cdot \frac{b_0}{\sqrt{b_1}}+(\sqrt{c_1}z+\frac{c_2}{2\sqrt{c_1}})\cdot \frac{c_0}{\sqrt{c_1}})^2 \leq (f(x,y,z)-(d-(\frac{a_2^2}{4a_1}+\frac{b_2^2}{4b_1}+\frac{c_2^2}{4c_1})))(\frac{a_0^2}{a_1}+\frac{b_0^2}{b_1}+\frac{c_0^2}{c_1}) $$

And from this , we obtain :- $$ f(x,y,z) \geq \frac{(l+\frac{a_2a_0}{2a_1}+\frac{b_0b_2}{2b_1}+\frac{c_0c_2}{2c_1})^2}{(\frac{a_0^2}{a_1}+\frac{b_0^2}{b_1}+\frac{c_0^2}{c_1})}+(d-(\frac{a_2^2}{4a_1}+\frac{b_2^2}{4b_1}+\frac{c_2^2}{4c_1})$$ Perhaps not the prettiest result , but certainly works when none of the denominators equate to $0$ . And in your case , on substituting the values , the formula happens to equal $\frac{31}{6}$ .