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.

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.
