不等式证明

Wed, 25 Feb 2026 21:44:13 +0800

题目描述

已知： $a > 0, b > 0$ 且 $a + b = 2$。

求证： $\sqrt{a+1} + \sqrt{b+1} \le 2\sqrt{2}$。

证明过程

方法一：利用最基础的代数运算性质

要证：$\sqrt{a+1}+\sqrt{b+1} \le 2\sqrt{2}$

即证：$(\sqrt{a+1}+\sqrt{b+1})^2 \le (2\sqrt{2})^2$

化简可得：$a+1+2\sqrt{(a+1)(b+1)}+b+1 \le 8$

整理可得：$a+b+2+2\sqrt{ab+a+b+1} \le 8$

代入$a+b=2$可得：$4+2\sqrt{ab+3} \le 8$

即$\sqrt{ab+3} \le 2$

两边平方可得：$ab \le 1$

代入$a+b=2$可得：$a(2-a) \le 1$

由于$a>0$且$b=2-a>0$，故$0<a<2$

由二次函数单调性可知当$a=1$时$a(2-a)$最大值为1

即$a(2-a) \le 1$成立，原不等式得证

方法二：利用柯西不等式 (Cauchy-Schwarz Inequality)

根据柯西不等式：$(x_1y_1 + x_2y_2)^2 \le (x_1^2 + x_2^2)(y_1^2 + y_2^2)$

令 $x_1 = \sqrt{a+1}, x_2 = \sqrt{b+1}$, $y_1 = 1, y_2 = 1$。

2025年寒假英语作文

Wed, 25 Feb 2026 14:50:38 +0800

应用文 - Should We Establish a Graffiti Wall?

题目

你校英文报 Campus Culture 栏目正在开展关于是否设立涂鸦墙（graffiti wall）讨论。请你写一篇短文投稿，内容包括：

1. 你的见解； 2. 你的建议。

范文

Recently, our school newspaper Campus Culture has initiated a discussion regarding（发起了一场关于…的讨论） the establishment of a graffiti wall on campus. Personally, I heartily endorse（由衷地支持） this proposal.

To begin with, a graffiti wall serves as an ideal vehicle（是…的理想载体） for students to unleash their creativity（释放他们的创造力） and vent their emotions（宣泄他们的情绪）. Given the grueling academic schedule（鉴于繁重的学业安排）, a designated space for artistic expression acts as a stress-reliever（起到了减压作用）, fostering individuality and imagination. Moreover, a well-curated graffiti wall can breathe new life into（为…注入新的活力） our campus culture, transforming dull corners into vibrant galleries. Rather than leaving random scribbles on desks, students are encouraged to showcase their talents in a more constructive and organized manner.

Ruoying Tan