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.
 A: 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*}
A: 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).
A: It is not true that $a^2\geq a$ for all $a\geq0$ (e.g. for $a=\frac12$), nor that $3b^2\geq b$ or $5c^2\geq c$.
Proof

 Because $c\geq a$, it suffices to prove $3a^2+3b^2+3c^2=1$. This follows from AM-QM: $\frac{a^2+b^2+c^2}3\geq\left(\frac{a+b+c}3\right)^2$.
 Or by using $3x^2\geq2x-\frac13$ (indeed, this is equivalent to $3(x-\frac13)^2\geq0$): $3a^2+3b^2+3c^2\geq2a+\frac13+2b+\frac13+2c+\frac13=1$.

