<s>[INST] 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.
 [/INST] 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.
</s>