44. Given an integer $n(n \geqslant 2)$. Find the largest constant $\lambda(n)$ such that: if the real number sequence $a_{0}, a_{1}, a_{2}, \cdots, a_{n}$ satisfies $0=a_{0} \leqslant a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$ and $a_{i} \geqslant \frac{1}{2}\left(a_{i-1}+a_{i+1}\right)(i=$ $1,2, \cdots, n)$, then $\left(\sum_{i=1}^{n} i a_{i}\right)^{2} \geqslant \lambda(n) \sum_{i=1}^{n} a_{i}^{2}$. (2009 China National Team Selection Test for IMO)
44. $\lambda(n)=\frac{n(n+1)^{2}}{4}$.

First, let $a_{1}=a_{2}=\cdots=a_{n}=1$. Then $\lambda(n) \leqslant \frac{n(n+1)^{2}}{4}$.
Next, we prove that for any sequence $a_{0}, a_{1}, a_{2}, \cdots, a_{n}$ satisfying the conditions, we have
$$\left(\sum_{i=1}^{n} i a_{i}\right)^{2} \geqslant \frac{n(n+1)^{2}}{4} \sum_{i=1}^{n} a_{i}^{2}$$

First, we prove: $a_{1} \geqslant \frac{a_{2}}{2} \geqslant \frac{a_{3}}{3} \geqslant \cdots \geqslant \frac{a_{n}}{n}$.
In fact, by the condition, we have $2 i a_{i} \geqslant i\left(a_{i-1}+a_{i+1}\right)$ for any $i(i=1,2, \cdots, n-1)$. For a given positive integer $l(1 \leqslant l \leqslant n-1)$, summing this inequality for $i(i=1,2, \cdots, l)$ gives $(l+$ 1) $a_{l} \geqslant l a_{l+1}$, i.e., $\frac{a_{l}}{l} \geqslant \frac{a_{l+1}}{l+1}$ for any $l(l=1,2, \cdots, n-1)$.

Next, we prove: for $i, j, k \in\{1,2, \cdots, n\}$, if $i>j$, then $\frac{2 i k^{2}}{i+k}>\frac{2 j k^{2}}{j+k}$.
In fact, the above inequality is equivalent to $2 i k^{2}(j+k)>2 j k^{2}(i+k) \Leftrightarrow(i-j) k^{3}>0$, which is clearly true. Now we prove inequality (1).
For $i, j(1 \leqslant i<j \leqslant n)$, we estimate the lower bound of $a_{i} a_{j}$.
By the previous result, we have $\frac{a_{i}}{i} \geqslant \frac{a_{j}}{j}$, i.e., $j a_{i}-i a_{j} \geqslant 0$.
Since $a_{i}-a_{j} \leqslant 0$, we have $\left(j a_{i}-i a_{j}\right)\left(a_{j}-a_{i}\right) \geqslant 0$, i.e.,
$$a_{i} a_{j} \geqslant \frac{i}{i+j} a_{j}^{2}+\frac{j}{i+j} a_{i}^{2}$$

Thus, we have
$$\begin{aligned}
\left(\sum_{i=1}^{n} i a_{i}\right)^{2}= & \sum_{i=1}^{n} i^{2} a_{i}^{2}+2 \sum_{1 \leqslant i<j \leqslant n} i j a_{i} a_{j} \geqslant \\
& \sum_{i=1}^{n} i^{2} a_{i}^{2}+2 \sum_{1 \leqslant i<j \leqslant n}\left(\frac{i^{2} j}{i+j} a_{j}^{2}+\frac{i j^{2}}{i+j} a_{i}^{2}\right)= \\
& \sum_{i=1}^{n}\left(a_{i}^{2} \sum_{k=1}^{n} \frac{2 i k^{2}}{i+k}\right)
\end{aligned}$$

Let $b_{i}=\sum_{k=1}^{n} \frac{2 i k^{2}}{i+k}$. By the previous proof, we have $b_{1} \leqslant b_{2} \leqslant \cdots \leqslant b_{n}$. Also, $a_{1}^{2} \leqslant a_{2}^{2} \leqslant a_{3}^{2} \leqslant \cdots \leqslant a_{n}^{2}$, so by Chebyshev's inequality, we have
$$\sum_{i=1}^{n} a_{i}^{2} b_{i} \geqslant \frac{1}{n}\left(\sum_{i=1}^{n} a_{i}^{2}\right)\left(\sum_{i=1}^{n} b_{i}\right)$$

Therefore,
$$\left(\sum_{i=1}^{n} i a_{i}\right)^{2} \geqslant \frac{1}{n}\left(\sum_{i=1}^{n} a_{i}^{2}\right)\left(\sum_{i=1}^{n} b_{i}\right)$$

And
$$\begin{aligned}
\sum_{i=1}^{n} b_{i}= & \sum_{i=1}^{n} \sum_{k=1}^{n} \frac{2 i k^{2}}{i+k}=\sum_{i=1}^{n} i^{2}+2 \sum_{1 \leqslant i<j \leqslant n}\left(\frac{i^{2} j}{i+j}+\frac{i j^{2}}{i+j}\right)= \\
& \sum_{i=1}^{n} i^{2}+2 i j=\left(\sum_{i=1}^{n} i\right)^{2}=\frac{n^{2}(n+1)^{2}}{4}
\end{aligned}$$

Thus, $\left(\sum_{i=1}^{n} i a_{i}\right)^{2} \geqslant \frac{n(n+1)^{2}}{4} \sum_{i=1}^{n} a_{i}^{2}$. Hence, inequality (1) is proved. In summary, $\lambda(n)=\frac{n(n+1)^{2}}{4}$.