<s>[INST] Prove that $\frac a{a+2b}+\frac bc+\frac{c^2}{\left(c+2a\right)^2 }\ge1$ 
(sqing) Let $a$, $b$, $c>0$. Prove that $\dfrac a{a+2b}+\dfrac bc+\dfrac{c^2}{\left(c+2a\right)^2 }\ge1$. Source (unsolved)

In this inequality, $b$ appears the least frequent, so I want to deal with it. I took $a$, $c$ as parameters and derived $b$, we have
\[\frac{\text d}{\text db}~\left(\dfrac a{a+2b}+\dfrac bc\right)=\frac1c-\frac{2a}{(a+2b)^2}.\tag1\label1\]
When $2c>a$, $\eqref1=0$ gives $b=\dfrac{\sqrt{2ac}-a}2$. When $2c\le a$, $\eqref1>0$ is true for all $b$, so $b\to0$ shall minimize the left side.
But plugging back makes the expressions complicated. What should I do?
 [/INST] Using Cauchy-Bunyakovsky-Schwarz inequality, we have
\begin{align*}
 &\frac{a}{a + 2b} + \frac{b}{c} + \frac{c^2}{(c + 2a)^2} - 1\\
 \ge{}& \frac{(a + b)^2}{a(a + 2b) + bc} + \frac{c^2}{(c + 2a)^2} - 1  \\
 ={}& \frac{(2ab - ac + bc)^2}{(a^2 + 2ab + bc)(c + 2a)^2}\\
 \ge{}& 0
\end{align*}
with equality if $a:b:c = 2:1:4$.
We are done.
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