Maximizing $ (4a-3b)^2+(5b-4c)^2+(3c-5a)^2$, such that $a^2+b^2+c^2=1 $ 
If $$a^2+b^2+c^2=1 $$here a,b,c are the real numbers then find the maximum value of $$ (4a-3b)^2+(5b-4c)^2+(3c-5a)^2$$

I tried to think with vectors, that is direction cosines of lines.
But then the expression is not getting simplified.
 A: Assume $\vec{A}=a\hat{i}+b\hat{j}+c\hat{k}$ and $\vec{B}=3\hat{i}+4\hat{j}+5\hat{k}$
$$\bigg|\vec{A}\times \vec{B}\bigg|^2=\bigg|\vec{A}\bigg|^2\bigg|\vec{B}\bigg|^2-\bigg(\vec{A}\cdot\vec{B}\bigg)^2$$
$$(4a-3b)^2+(5b-4c)^2+(3c-5a)^2=50-(3a+4b+5c)^2\leq 50$$
A: Using the method of Lagrange multipliers, consider
$$F= (4a-3b)^2+(5b-4c)^2+(3c-5a)^2+\lambda(a^2+b^2+c^2-1) \tag 1$$ Computing the partial derivatives
$$\frac{\partial F}{\partial a}=8 (4 a-3 b)-10 (3 c-5 a)+2 a \lambda \tag 2$$
$$\frac{\partial F}{\partial b}=-6 (4 a-3 b)+10 (5 b-4 c)+2 b \lambda\tag 3$$
$$\frac{\partial F}{\partial c}=6 (3 c-5 a)-8 (5 b-4 c)+2 c \lambda\tag 4$$
$$\frac{\partial F}{\partial \lambda}=a^2+b^2+c^2-1\tag 5$$
From $(2)$, $b=\frac{1}{12} (a \lambda +41 a-15 c)$; plug in $(3)$ to get $c=\frac{1}{15} (a \lambda +25 a)$ which makes $b=\frac{4 }{3}a$; plug $b$ and $c$ in $(4)$ to get
$$\frac{2}{15} a \lambda  (\lambda +50)=0 \tag 6$$ So, we have three cases : $a=0$ , $\lambda=0$, $\lambda=-50$. 
The first case $a=0$ can be discarded since it would make $a=b=c=0$ which does not satisfy the constraint.
The second case $\lambda=0$ would make $c=\frac{5 }{3}a$ which, in turn, would make $\frac{50 }{9}a^2=1$ from the constraint and then the value of the expression to maximize would just be $0$.
So, what is left is the case $\lambda=-50$ which makes $c=-\frac{5 }{3}a$ which gives again $\frac{50 }{9}a^2=1$ from the constraint. The expression to maximize if then $\frac{a^2 \lambda ^2}{9}$ with $\lambda=-50$ and $\frac{50 }{9}a^2=1$ which then gives a maximum vzlue of $50$.
I let you finishing what could be required.
A: We need to find a minimal value of $k$ for which the following inequality is true  for all reals $a$, $b$ and $c$.
$$(4a-3b)^2+(5b-4c)^2+(3c-5a)^2\leq k(a^2+b^2+c^2)$$ or
$$(k-41)a^2+(k-34)b^2+(k-25)c^2+24ab+40bc+30ac\geq0$$ or
$$(k-41)a^2+6(4b+5c)a+(k-34)b^2+40bc+(k-25)c^2\geq0,$$ for which we need
$$k>41$$ and
$$9(4b+5c)^2-(k-41)((k-34)b^2+40bc+(k-25)c^2)\leq0$$ or
$$(k^2-75k+1250)b^2+(k^2-66k+800)c^2+40(k-50)bc\geq0$$ or
$$(k-50)((k-25)b^2+(k-16)c^2+40bc)\geq0.$$
But $$(k-25)b^2+(k-16)c^2+40bc\geq16b^2+25c^2+40bc=(4b+5c)^2\geq0,$$ which gives $$k\geq50.$$
For $k=50$ we obtain $$9a^2+6(4b+5c)a+(4b+5c)^2\geq0$$ or
$$(3a+4b+5c)^2\geq0,$$ which gives that the equality occurs for $$3a+4b+5c=0$$ and $$a^2+b^2+c^2=1,$$
which says that $50$ is a maximal value.
A: Use the technique of Lagrange multipliers to solve. 
Essentially, if $$f(a,b,c) = a^2 + b^2 + c^2,$$ $$g(a,b,c) = (4a-3b)^2+(5b-4c)^2+(3c-5a)^2,$$ then find the values $(a,b,c)$ such that $\vec \nabla f$ is parallel to $\vec \nabla g$. 
Then, by a bit of testing on your solutions, select the one that satisfies $f(a,b,c) = 1$ and is a maximum of $g(a,b,c)$ (as opposed to a minimum or saddle point, for instance).
A: try opening the brackets.
square terms with coeeficents combine them
replace $$ a^2+b^2$$ type terms with $$c^2$$.
you have 
$$ 50-(3a+4b+5c)^2$$
