let $\{b_{n}\}$ such that $b^2_{n}=b_{n+1}(1+b^2_{n})$, and $$T_{n}=\sum_{k=1}^{n}\dfrac{(-1)^k(k+2)}{(k+1)^2}b_{k}$$ show that
$T_{n}<\dfrac{5}{4}|b_{1}|$
my idea $b^2_{n}=b_{n+1}(1+b^2_{n})$ then $b_{n}>0,n\ge 2$
and $$b^2_{n}\ge 2b_{n+1}b_{n},\Longleftrightarrow b_{n+1}\le \dfrac{1}{2}b_{n}$$ then $$b_{n}\le \dfrac{1}{2^{n-2}}b_{2}=\dfrac{1}{2^{n-2}}\dfrac{b^2_{1}}{1+b^2_{1}},n\ge 2$$
But is follow very ugly. Thank you veryone
Suppose $b_1\ne 0$. (If $b_1=0$, then $b_n=0$ for all $n$ and the inequality does not hold.) And I also assume $n\ge 1$ in $b_n^2=b_{n+1}(1+b_n^2)$.
Lemma. $0<b_{n+1}\leq b_n/2$ for $n\ge 2$, and $0<b_2\leq |b_1|/2$.
Proof of lemma. (From the OP's argument.) $b_{n+1}=\frac{b_n^2}{1+b_n^2}>0$ for $n\ge 1$. Using the fact that $1+b_n^2\ge 2|b_n|$, we find $|b_n|^2=b_n^2=b_{n+1}(1+b_n^2)\ge 2b_{n+1}|b_n|$, hence $|b_{n}|\ge 2b_{n+1}$.
Proof. The $(2l-1)$-th and $2l$-th terms of $T_n$ ($l\ge 1$) are \begin{align*} -\frac{2l+1}{(2l)^2}b_{2l-1}+\frac{2l+2}{(2l+1)^2}b_{2l}. \end{align*}
For $l\ge 2$, this is negative, because $\dfrac{2l+1}{(2l)^2}\geq \dfrac{2l+2}{(2l+1)^2}>0$ ($\because\dfrac{k+2}{(k+1)^2}=\dfrac1{k+1}+\dfrac1{(k+1)^2}$ is a decreasing function of $k\ge 1$) and $0<b_{2l}\leq b_{2l-1}/2$ (from the lemma).
For $l=1$, this is equal to \begin{align*} -\frac34b_1+\frac49b_2 &\leq\frac34|b_1|+\frac49\cdot\frac12|b_1|\\ &=\frac{35}{36}|b_1| \end{align*}
If $n$ is even, the above argument suggests $T_n\le\frac{35}{36}|b_1|<\frac54|b_1|$. If $n$ is odd ($\ge 3$), the last term remains, which is negative, so still $T_n\le\frac{35}{36}|b_1|<\frac54|b_1|$.
