Problem 1524

Let $P_{n}=a^{n}+a^{n-1} b+a^{n-2} b^{2}+\cdots+a b^{n-1}+b^{*}, Q_{n}=$ $\left(\frac{a+b}{2}\right)^{n}$. Then prove the original inequality is equivalent to proving $P_{n} \geqslant(n+1) Q_{n}$.

Since $(a+b) P_{n-1}+a^{n}+b^{n}=2 P_{n}$, i.e., $P_{n}=\frac{a+b}{2} P_{n-1}+\frac{a^{n}+b^{n}}{2}$. By the mean value inequality, we get $P_{n} \geqslant \frac{a+b}{2} P_{n-1}+\left(\frac{a+b}{2}\right)^{n}$, so $\frac{P_{n}}{Q_{n}} \geqslant \frac{P_{n-1}}{Q_{n-1}}+1$. Therefore,
$$\begin{array}{l} \text { it follows that } \frac{P_{n}}{Q_{n}} \geqslant \frac{P_{n-1}}{Q_{n-1}}+1 \geqslant\left(\frac{P_{n-2}}{Q_{n-2}}+1\right)+1=\frac{P_{n-2}}{Q_{n-2}}+2 \geqslant \cdots \geqslant \frac{P_{1}}{Q_{1}}+(n-1) \\ =2+(n-1)=n+1 . \end{array}$$

Thus, $\frac{P_{n}}{Q_{n}} \geqslant n+1$, and the original inequality is proved.

Prove by mathematical induction:
(1) When $n=2$, since $a^{2}+b^{2} \geqslant 2 a b$, adding $3\left(a^{2}+b^{2}\right)+$ $4 a b$ to both sides gives $4\left(a^{2}+b^{2}+a b\right) \geqslant 3(a+b)^{2}$, i.e., $\frac{a^{2}+b^{2}+a b}{3} \geqslant\left(\frac{a+b}{2}\right)^{2}$, the inequality holds, with equality if and only if $a=b$.
(2) Assume the inequality holds for $n=k-1$, then for $n=k$, because
$$a^{k}+b^{k} \geqslant a^{k-i} b^{i}+a^{i} b^{k-i}(i=1,2, \cdots, k-1)$$

Adding these inequalities gives
$$(k-1)\left(a^{k}+b^{k}\right) \geqslant 2\left(a^{k-1} b+a^{k-2} b^{2}+\cdots+a b^{k-1}\right),$$

Adding $(k+1)\left(a^{k}+b^{k}\right)+2 k\left(a^{k-1} b+a^{k-2} b^{2}+\cdots+a b^{k-1}\right)$ to both sides gives
$$\begin{aligned} & 2 k\left(a^{k}+a^{k-1} b+a^{k-2} b^{2}+\cdots+a b^{k-1}+b^{k}\right) \\ \geqslant & {\left[a^{k}+2\left(a^{k-1} b+a^{k-2} b^{2}+\cdots+a b^{k-1}\right)+b^{k}\right] } \end{aligned}$$

Dividing both sides by $2 k(k+1)$ gives
$$\begin{aligned} & \frac{a^{k}+a^{k-1} b+a^{k-2} b^{2}+\cdots+a b^{k-1}+b^{k}}{k+1} \\ \geqslant & \frac{a^{k}+2\left(a^{k-1} b+a^{k-2} b^{2}+\cdots+a b^{k-1}\right)+b^{k}}{2 k} \\ = & \left(\frac{a^{k-1}+a^{k-1} b^{2}+\cdots+a b^{k-2}+b^{k-1}}{k}\right)\left(\frac{a+b}{2}\right) \\ \geqslant & \left(\frac{a+b}{2}\right)^{k-1}\left(\frac{a+b}{2}\right)=\left(\frac{a+b}{2}\right)^{k} \end{aligned}$$

Equality holds if and only if $a=b$, i.e., the proposition holds for $n=k$.
By (1) and (2), the inequality holds for $n \in \mathbf{N}^{*}$ and $n \geqslant 2$.