Let $0 \le a \le b \le c$ and $a+b+c=1$. Show that $a^2+3b^2+5c^2 \ge 1$ 
Let   $0 \le a \le b \le c$ and $a+b+c=1$. Show that $a^2+3b^2+5c^2
 \ge 1$.

My solution: since $a+b+c=1$ we have to show that $a^2+3b^2+5c^2\ge1=a+b+c$
Since $a,b,c \ge 0 $  the inequality is true given that every term on the left hand side of the inequality is greater or equal to the corresponding term on the right.
However I am not sure if I am reasoning correctly, as the hint from my book seems to depict the problem in a harder way than I am ,as it suggests to square the expression $a+b+c=1$ and so on...
So my question is wheter I am overlooking some detail in the problem which makes my solution inadequate.

By means of rearrangement inequality, it can be shown that, is $a_1\le a_2\le\cdots\le a_n$ and $b_1\le b_2\le\cdots\le b_n$, then $\dfrac{a_1b_1+a_2b_2+\cdots+a_nb_n}{n}\ge \dfrac{a_1+a_2+\cdots+a_n}{n}\dfrac{b_1+b_2+\cdots+b_n}{n}$
Then, $\dfrac{a^2+3b^2+5c^2}{3}\ge\dfrac{a^2+b^2+c^2}{3}\dfrac{1+3+5}{3}$. Thus $a^2+3b^2+5c^2\ge 3(a^2+b^2+c^2)\ge 3\dfrac{(a+b+c)^2}{3}=1$ (last follows from Jensen's inequality).
