Can we AM-GM it?

Observe the following:

$\begin{aligned} a^2+3b^2+5c^2 &= a^2+b^2+c^2+2b^2+2c^2+2c^2\\ &\leq a^2+b^2+c^2+2ab+2ac+2bc\\ &=(a+b+c)^2\\ &=441 \end{aligned}$

Thus, $a^2+3b^2+5c^2\leq 441$ . Equality occurs in three different cases which we shall also explore. We have

$\begin{aligned} a^2+3b^2+5c^2&=a^2+b^2+c^2+2ab+2bc+2ca\\ b^2-ab+c^2-bc+c^2-ca&=0\\ b(b-a)+c(c-b)+c(c-a)&=0 \end{aligned}$

But we have that $a \geq b \geq c \geq 0$ . Thus, we have $b(b-a) \leq 0$ , $c(c-b) \leq 0$ and $c(c-a) \leq 0$ . From this, we must have all the terms in the sum to equal to 0. Thus, we have either $a=b=c=7$ or $a=21$ and $b=c=0$ or $a=b=\frac{21}{2}$ and $c=0$ . Thus, we have found all solutions for equality.

Based on the condition, we can prove that $f(a,b,c)\leq f(a,a,c)$ or $a^2+3b^2+5c^2\leq 4a^2+5c^2$ Since $a\geq b\geq c\geq 0$ and $a+b+c=21$ . Considering $f(a,a,c)$ , the condition becomes $2a+c=121 \Rightarrow c=21-2a\geq 0\Rightarrow a\in [7;\frac{21}{2}]$ $\Rightarrow f(a,a,c)=4a^2+5(21-2a)^2=24a^2-420a+2205=f(a)_1$ We see that $f(a)_{1_{min}}= \frac{735}{2}$ when $a=\frac{35}{4}\in [7;\frac{21}{2}]$ so now we can find the maximum by comparing $f(7)$ with $f\big(\frac{21}{2}\big)$ which turns out to be equal to $441$ , so we have 2 equality cases $(a,b,c)=(7,7,7);\big(\frac{21}{2},\frac{21}{2},0\big)$

Now I'll try to mix these variables in another way, that means $f(a,b,c)\leq f(a,b,b)$ . Now the condition becomes $a+2b=21 \Rightarrow 2b=21-a\geq 0$ , therefore we get the interval $a\in [7;21]$ . Now we have $f(a,b,b)=a^2+8b^2=a^2+2(21-a)^2=3a^2-84a+882=f(a)_2$ We see that $f(a)_{2_{min}}= 294$ when $a=14 \in[7;21]$ , so the max'll be found by comparision between $f(7)$ and $f(21)$ , which also give us the same result as before. Now we have the third equality cases $(a,b,c)=(21;0;0)$

From 2 methods of mixing variables, we have $a^2+3b^2+5c^2\leq 441$ and 3 equality cases $(a,b,c)=(21;0;0),\big(\frac{21}{2};\frac{21}{2};0),(7;7;7)$ . @Pi Han Goh, can you check it out