An algebra problem by Vaibhav Prasad

$a^2 +b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ac) = 36 - 2(ab+bc+ac) \ (1)$ Now we must maximise 2(ab +bc +ac) in order to minimise our expression. So using the Cauchy Schwarz inequality : $(ab+ac+bc)^2 \leq\ (a^2+b^2+c^2)(a^2+b^2+c^2)$ $\Rightarrow\ max(ab+ac+bc) = a^2 +b^2+c^2$ Now we can substitute this into (1) to give: $36 = 3(a^2 + b^2 + c^2) \therefore\ min(a^2+b^2+c^2) = \boxed{12}$ $\large when \ a=b=c= 2$

Every solution here is very algebraic. There is a geometric way to see this. The conditions imply that (a,b,c) must lie somewhere on the triangle spanned by (6,0,0),(0,6,0) and (0,0,6). The expression $a^2+b^2+c^2$ is the square of the norm of the vector (a,b,c). So we are looking for the vector of least length. By "picture" or by symmetry considerations, we see that this occurs on the midpoint of the triangle, or in other words, when $a=b=c$ , which can only happen when $a=b=c=2$ . Hence the answer is $4 \cdot 3 = 12$ .

By Cauchy Schwarz inequality ,

$a^2 + b^2 + c^2 \geq \dfrac{(a+b+c)^2}{1+1+1}$

$\implies$ $a^2 + b^2 + c^2 \geq 12$ .

$a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)$ and $ab+ac+bc \leq a^2+b^2+c^2$ (by AM-GM inequality,Cauchy or simple rearrengment) , we have

$a^2+b^2+c^2 \geq (a+b+c)^2-2(a^2+b^2+c^2)$

$3(a^2+b^2+c^2) \geq (a+b+c)^2 = 36$

$a^2+b^2+c^2 \geq 12$

Equality occurs when $a=b=c = 2$

I used LaGrange :v

Its quite simple:

restriction : a+b+c=6

derivate in a,b and c both sides and lambda.

λ:2a

λ:2b

λ:2c

So a=b=c, aplying in the restriction

a+a+a=6

3a=6

a=2=b=c

(2)^2+(2)^2+(2)^2=4+4+4=12

1 + 2 + 3 = 6 & 1^2 + 2^2 + 3^2 = 14
1 + 1 + 4 = 6 & 1^2 +1^2 + 4^2 = 18
2 + 2 + 2 = 6 & 2^2 + 2^2 + 2^2 = 12 Ans.
Note: as per given condition, values of a, b, c should be minimum and here less than 2 is not possible.

There is a interesting geometric solution:

1. a+b+c=6 is a plane. The equation is $1.a + 1.b + 1.c - 6 = 0$
2. a, b, c are axes.
3. Thus, $a^2 + b^2 + c^2$ is is the square of the norm of the vector that goes from (0,0,0) to the plane.
4. A minimal $a^2 + b^2 + c^2$ is square of the norm of the perpendicular vector that goes from (0,0,0) to the plane.
5. Thus, a minimal $a^2 + b^2 + c^2$ is the square of distance of plane $1.a + 1.b + 1.c - 6 = 0$ to (0,0,0).
6. The formula for a distance d is: $d = \frac{ |1.0 + 1.0 + 1.0 - 6|}{ \sqrt{1^2 + 1^2 + 1^2}}$
7. Thus, $d = \frac{6}{\sqrt{3}}$ and $d^2 = a^2 + b^2 + c^2 = 12$

Isn't it soooo obvious that the Minimum value comes with the lesser square possible? In other words:

If you need to satisfy a + b + c = 6 with numbers wich squares are the lesser, it obviously will be the middle number as possible (in this case 2 for the three, cause in the others ways you can take 1 or 0 in one incognite, but have to use numbers much biggers in one of the others to compensate: example 0, 1, 5; 0, 0, 6; and so on - even 1, 2, 3 the square of 3 is much bigger than the square of 2!! i hope i could express my thinking). Isn't it logical?

Why all those counts?

Anyone else who used Lagrange Multiplier? :)

Using the extended version of the AM-GM Inequality, the QM-AM-GM-HM Inequality, we know that (using the first two inequalities)

$\displaystyle \sqrt{\dfrac{\sum\limits_{i=1}^{n} {(x_i)}^{2} }{n}} \geq \dfrac{\sum\limits_{i=1}^{n} x_i}{n}$

For $n = 3$ , this simplifies to

$\displaystyle \sqrt{\dfrac{a^2 + b^2 + c^2}{3}} \geq \dfrac{a+b+c}{3}$

Since $a+b+c=6$ ,

$\displaystyle \sqrt{\dfrac{a^2 + b^2 + c^2}{3}} \geq \dfrac{6}{3} = 2$

$\displaystyle \implies \dfrac{a^2 + b^2 + c^2}{3} \geq 2^2$

$\displaystyle \implies a^2 + b^2 + c^2 \geq 12$

Therefore, the minimum value of $a^2 + b^2 + c^2 = \boxed{12}$ with equality occuring when $a=b=c=2$ .

When I think a + b + c = 6, the minimum value of a,b and c is 2 that is a=b=c=2, therefore substituting the values, 4+4+4=12

All you need to really do is see what is the smallest integers that can satisfy the second equation. That would be... a = 2 b = 2 c = 2 2 + 2 + 2 = 6 2^2 + 2^2 + 2^2 = 4 + 4 + 4 = 8 + 4 = 12

I used Lagrange multipliers to solve the problem, went rather smoothly.

Guys I see this in a much more simpler light. All one has to do is equally distribute the sum amongst each of the numbers to get the least value for the sum of their squares. This should also work for 4, 5, 6 or n numbers. Try it. In fact this should work for the sum of the cubes or the 4th powers and so on. For example what is the minimum value of a^2 + b^2 + c^2 + d^2 when a + b + c + d = 8. It is true when they are equally distributed and a = b = c = d = 2. Intuitive solution is better and easier, eh?!! ;-).

(a^2+b^2+c^2)'=2a+2b+2c=2(a+b+c)

Max value for a+b+c=6 so,

2*6=12

(I don't see this solution so I must be wrong somewhere..)

Using the method of Lagrange Multipliers, this problem can be easily approached. First let's define the function to be minimized as $f(a,b,c)=a^{2}+b^{2}+c^{2}$ and the constrain function to be $g(a,b,c)=a+b+c=6$ By Lagrange Multipliers the gradient of the minimized function is proportional to the constrain function, thus $\nabla f(a,b,c)=\lambda \nabla g(a,b,c)$ the gradient of the minimized function is $\nabla f(a,b,c)= 2<a,b,c>$ and the gradient of the constrained function is $\nabla g(a,b,c)= \lambda<1,1,1>$

We are now left with three equations - $2x=\lambda$ - $2y=\lambda$ - $2z=\lambda$ which implies that $x=y=z$ . Because of this we can now substitute y and z into our constraint equation yielding the equation

$x+x+x=6 \Rightarrow x=2$

Now that we finally have revealed the values of x, y, and z we can substitute it into our function to be minimized leaving us with

$2^{2}+2^{2}+2^{2}=\boxed{12}$

Without knowing much of hardcore math, such as the Cauchy Schwarz inequality, I took a more intuitive and empirical approach.

Let's assume $min(a^{2} + b^{2} + c^{2})$ happens for $a = b = c = 2$ . Now, if that's wrong, I should be able to obtain a lower value for $a^{2} + b^{2} + c^{2}$ by locking one of the variables, let's say $a$ , and raising another one, let's say $b$ , which would naturally drop $c's$ value equivalently, since we have $a + b + c = 6$ .

But we know that the exponential function of power of 2 will grow higher than it will diminish for a giving $epsilon$ if we have a positive base.

Therefore any raise in b and equivalent drop in $c$ would cause $a^{2} + b^{2} + c^{2}$ to raise its value.

Can't relate to the discussion 😅😅😅

The minimum value for $a^2 + b^2 + c^2$ is also the minimum value for $\sqrt{a^2 + b^2 + c^2}$ . The solution is where the sphere of radius $\sqrt{a^2 + b^2 + c^2}$ touches the plane $a + b + c =6$ . Since a,b,c contribute equally to the linear equation then they must be equal to 2, which gives $a^2 + b^2 + c^2 = 12$ .

You will get the min value when all the terms (a, b, c) are minimum. And that happens when a=b=c =6

By Cauchy-Schwarz,

$36=(a+b+c)^{2} \leq (a^{2}+b^{2}+c^{2})(1^{2}+1^{2}+1^{2})=3 (a^{2} +b^{2}+c^{2})$

Thus the answer is at least 12. But $a=b=c=2$ gives the value of 12. Thus 12 is the answer.

A=B=C=2 GIVES THE MINIMUNM VALUE THE CONCEPT COMES FROM AM>=GM

It is known that : $a^2 + b^2 + c^2 \geq\ ab + bc + ca \ (1)$ The equal sign is for a = b = c. This comes from: $(a - b)^2 \geq 0$ and 2 similars to, for b with c and c with a. Equal, evidently obtained, for a = b, b = c and respectively c = a. Summing the 3 inequalities & simplifying by 2 we obtain the (1) relation . We have: $(a+b+c)^2 = 6^2\rightarrow \ a^2+b^2+c^2+2(ab+bc+ca) = 36 \ (2)$ Substituting (1) in (2) we obtain: $a^2+b^2+c^2+2(a^2+b^2+c^2)\geq 36$ & simplifying by 3 we obtain the answer: $\rightarrow \ a^2+b^2+c^2 \geq 12$ Equal is obtained for: a = b = c

All the three numbers are greater than zero,and their sum is 6.So,the three nunumbers can be 1,2and 3 or 2,2 and 2.In the second case,we get the least number i.e 12! and voila I got the answer with no algebra.....: P

Let u =(1, 1, 1), v = (a, b, c) be two vectors, the acute angle between them be x

The scalar product of u, v = u . v = a + b + c = 6

Norm of u = ||u|| = square root of 3

Norm of v = ||v|| = square root of (a^2 + b^2 + c^2)

cos x = ( u . v )/||u||.||v|| = 6/square root of 3(a^2 + b^2 + c^2)

cos x is less than or equal to 1

cos x is more than or equal to 0 ...........(x is acute)

Then

a^2 + b^2 + c^2 is more than or equal 12

Then

The minimum value of a^2 + b^2 + c^2 is 12

a+b+c=6
6/3 =2 2 2+2 2+2*2=4+4+4 =12 a,b,c are three numbers so if divided by total value get the variables value.

Using the jensen's inequality for the function X--> X^2 we conclude immediately

2 + 2 + 2 = 6 & 2^2 + 2^2 + 2^2 = 12 Ans

can you plz explain the title @Vaibhav Prasad

Guess and check 1+2+3=6 (1^{2})+(2^{2})+(3{^2})=14 2+2+2+6 3*(2{^2})=12 minimum value=12

Why not use Python ?

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from itertools import product
tests = xrange(1, 5)
smallest = 1000000
for a,b,c in product(tests,repeat=3):
    if a+b+c == 6:
        if a*a+b*b+c*c < smallest:
            smallest = a*a+b*b+c*c
print "Answer:", smallest

Since the AM of nth powers is greater than nth power of AM of the terms... Therefore {a^2 + b^2 +c^2}/3 > 4 and hence the result..