Warm up problem 18: Find the minimum... but it's not symmetric.

using Cauchy inequality :

$\quad { ((a)\times (1)\quad +\quad (\sqrt { 2 } b)\times (\cfrac { 1 }{ \sqrt { 2 } } )\quad +\quad (c)\times (1) })^{ 2 }\quad \le \quad ({ a }^{ 2 }+{ 2b }^{ 2 }+{ c }^{ 2 })({ 1 }^{ 2 }+{ (\cfrac { 1 }{ \sqrt { 2 } } ) }^{ 2 }+{ 1 }^{ 2 })\\ \\ { ({ a }^{ 2 }+{ 2b }^{ 2 }+{ c }^{ 2 }) }_{ min }\quad =\quad 2/5\quad =\quad 0.4$ .

Using Cauchy-Schwartz inequality he have:

$\left( 1^2 + (\frac {1}{2})^2 + (\frac{1}{2})^2+1^2\right) \left( a^2 + b^2 + b^2 + c^2 \right) \ge \left( a+\frac {b}{2}+\frac{b}{2}+c\right)^2$

$\Rightarrow \frac {5}{2} (a^2 + 2b^2+c^2) \ge (a+b+c)^2 = 1$

$\Rightarrow a^2 + 2b^2+c^2 \ge \boxed {\frac{2}{5}}$

For sake of variety, how about some Lagrange multipliers? We wish to minimize $f(a,b,c) = a^{2} + 2b^{2} + c^{2}$ given the restraint $g(a,b,c) = a + b + c - 1 = 0$ .

Then the Lagrange equations $\frac{\partial}{\partial (a,b,c)} f(a,b,c) = \lambda \frac{\partial}{\partial (a,b,c)} g(a,b,c)$ are

$2a = \lambda, 4b = \lambda, 2c = \lambda \Longrightarrow a = \frac{\lambda}{2}, b = \frac{\lambda}{4}, c = \frac{\lambda}{2}$ .

Plugging these into the constraint gives us that $\frac{5 \lambda}{4} = 1 \Longrightarrow \lambda = \frac{4}{5}$ .

Thus $a = c = \frac{2}{5}$ and $b = \frac{1}{5}$ , giving us

$f(\frac{2}{5},\frac{1}{5},\frac{2}{5}) = \frac{4}{25} + 2*\frac{1}{25} + \frac{4}{25} = \frac{10}{25} = \frac{2}{5} = \boxed{0.4}$ as the minimum.

(Note that, for example, if $a = b = c = \frac{1}{3}$ we would have $a^{2} + 2b^{2} + c^{2} = \frac{4}{9} \gt \frac{2}{5}$ , so we can conclude that the solution found above is indeed the minimum. Also note that by this method we can drop the restriction that $a,b,c$ be positive.)

It is symmetric about a and c , Thus putting a=c=p

we have 2p+b=1

and

$2{ p }^{ 2 }+2{ b }^{ 2 }=1\quad \\$

solving above equations yield the quadratic $10{ p }^{ 2 }-8p+2\\$

which gives a minimum at p=0.4 which is 0.4 itself

Applying Cauchy-Schwarz's Inequality, we have $(a^2+2b^2+c^2)(1+\frac{1}{2}+1) \ge (a+b+c)^2 =1$ Thus, $(a^2+2b^2+c^2) \ge \frac{1}{1+\frac{1}{2}+1}=\frac{2}{5}$ Therefore the minimum possible value is $\frac{2}{5}=0.4$

A physicist's full proof:

We can see that $b$ carries a lot more weight in the final sum $a^2+2b^2+c^2$ . Seeing this and also that $a$ and $c$ must be symmetric, we can guess that the minimum value of the second sum happens when $a$ and $c$ are equal, $b$ is half the value of those two, and their sum is equal to $1$ . This happens when

$a=c=\frac{2}{5}$

$b=\frac{1}{5}$

Which would give us $a^2+2b^2+c^2=\frac{10}{25}=\boxed{0.4}$ .

Let $k=a+c=1-b$ , ie. $c=k-a$ , $b=1-k$

${ a }^{ 2 }+2 { b }^{ 2 }+{ c }^{ 2 }\\ ={ a }^{ 2 }+{ c }^{ 2 }+2{ b }^{ 2 }\\ ={ a }^{ 2 } +{ (k-a) }^{ 2 }+2(1-k)^{ 2 }\\ =2{ a }^{ 2 }-2ka+{ a }^{ 2 }+2 { k }^{ 2 }-4k+2\\ =2(a-\frac { 1 }{ 2 } k)^{ 2 }+\frac { 5 }{ 2 } (k-\frac { 4 }{ 5 } )^{ 2 }+\frac { 2 }{ 5 }$

Hence, when $c=a=\frac { 1 }{ 2 } k$ , $k=\frac { 4 }{ 5 }$ , ie. $b=1-\frac { 4 }{ 5 }=\frac { 1 }{ 5 }$

${ a }^{ 2 } + 2 { b }^{ 2 } + { c }^{ 2 }$ has minimum value of $\frac { 2 }{ 5 }$

as someone mentioned, I also figured out that this would be symmetric about a and c.

since a+b+c =1 and a^2 + 2 b^2 + c^2 both are having equal weight of a and c, I assumed a=c=k

now

it's a simple differentiation problem given 2a+b =1 minimize 2 a^2 + 2 b^2, simply by taking derivative of b w.r.t a.

Used Titu's Lemma where
$a_1=a$ ,
$a_2=b$ ,
$a_3=c$ ,
$b_1=1$ ,
$b_2=\frac{1}{2}$ and
$b_3=1$

Computing is generally applicable for many types of questions:

From 0.4444 to 0.4, answer is 0.4 with a = 2/ 5, b = 1/ 5 and c = 2/ 5.

PROGRAM Search;

USES CRT;

VAR a, b, c: LONGINT; m, y: EXTENDED;

BEGIN CLRSCR; m:=4/ 9;

 FOR a:=1 TO 1000 DO
     FOR b:=1 TO 1000 DO
     BEGIN
          c:=1000-a-b;
          y:=SQR(0.001*a)+2.0*SQR(0.001*b)+SQR(0.001*c);
          IF (c>0) AND (y<m) THEN
          BEGIN
             m:=y;
             WRITELN(m:1:4, ' ', 0.001*a:1:2, ' ', 0.001*b:1:2, ' ', 0.001*c:1:2);
          END
     END

END.

Since a = c ought to be true after intuition from above,

a + b + a = 1 and a^2 + 2 b^2 + a^2 need for a minimum,

2 a^2 + 2 (1 - 2 a)^2 = 2 [5 (a - 2/ 5 )^2 + 1/ 5] means a = c = 2/ 5 while b = 1 - 4/ 5 = 1/ 5 for 2 [1/ 5] = 0.4 {Answer}

We can see that (b) carries a lot more weight in the final sum a^2 + 2 b^2 + c^2 .

Seeing this and also that (a) and (c) must be symmetric, we can guess that the minimum value of the second sum happens when (a) and (c) are equal, (b) is half the value of those two, and their sum is equal to 1 . This happens when:

a = c = 2/5

b = 1/5

Which would give us a^2 + 2 b^2 + c^2 = (2/5)^2 + 2 (1/5)^2 + (2/5)^2 = 4/10

= 2/5 = 0.4

Well i'm not that good with inequalities so I will solve this problem without complicated inequalities or theorems

First, according to QM - AM
$a^{2} + c^{2}$ obtain minimum at $a = c$

We get $2a + b = 1$ and want to minimize $2a^{2} + 2b^{2}$

Hint : find a circle centered at $(0,0)$ with the smallest radius $r$ but still intersect the line $2a + b = 1$

The answer to this problem is the value of $2r^{2}$

you can prefer the solution using lagrange multiplier