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.

Based on the inequality 
\begin{align*}
\frac{x^2}{a} + \frac{y^2}{b} +\frac{z^2}{c} \ge \frac{(x+y+z)^2}{a+b+c}.
\end{align*}
we have shown in This Question, for any positive $x$, $y$, $z$, $a$, $b$, and $c$, we have that
\begin{align*}
a^2 + 3b^2 + 5c^2 &\ge 3a^2 + 3 b^2+3c^2\\
&\ge \frac{(a+b+c)^2}{\frac{1}{3} + \frac{1}{3}+\frac{1}{3}} =1.
\end{align*}
