不等式技巧

(2017 USAMO) Find the minimum possible value of $$\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4}$$ given that $a$, $b$, $c$, $d$ are nonnegative real numbers such that $a+b+c+d=4$.

Proposed by Titu Andreescu

I claim $\sum_{cyc} \dfrac{a}{b^3+4} \ge \dfrac{2}{3}$.
Subtracting $\dfrac{a+b+c+d}{4} = 1$ from both sides, we get $$-\dfrac{1}{4}\sum_{cyc} \dfrac{ab^3}{b^3+4} \ge -\dfrac{1}{3}\iff \sum_{cyc} \dfrac{ab^3}{b^3+4} \le \dfrac{4}{3}$$ But since $\dfrac{b^3}{b^3+4} \le \dfrac{b}{3}\impliedby b(b+1)(b-2)^2 \ge 0$ true for $b\ge 0$, it suffices to prove $$\sum_{cyc} \dfrac{ab}{3} \le \dfrac{4}{3} \iff (a+b+c+d)^2 \ge 4(ab+bc+cd+da)$$ $$\iff (a-b+c-d)^2\ge 0$$ which is true.
Equality case at $(a,b,c,d)=(2,2,0,0)$ and cyclic variants.

 

... and it really happened. Fortunately, it's solved by the tangent line trick.

We observe the miraculous identity $$\frac{1}{b^3+4} \ge \frac14 - \frac{b}{12}$$ since $12-(3-b)(b^3+4) = b(b+1)(b-2)^2 \ge 0$. Moreover, $$ab+bc+cd+da = (a+c)(b+d) \le \left( \frac{(a+c)+(b+d)}{2} \right)^2 = 4.$$ Thus $$\sum_{\text{cyc}} \frac{a}{b^3+4} \ge \frac{a+b+c+d}{4} - \frac{ab+bc+cd+da}{12} \ge 1 - \frac13 = \frac23.$$ This minimum $\frac23$ is achieved at $(a,b,c,d) = (2,2,0,0)$ and permutations.

 

(2007年第6届中国女子数学奥林匹克（CGMO）)设整数$n\,(n>3)$,非负实数$a_1,a_2,\cdots,a_n$满足$a_1+a_2+\cdots+a_n=2$.求 $$\frac{a_1}{a_2^2 + 1}+ \frac{a_2}{a^2_3 + 1}+ \cdots + \frac{a_n}{a^2_1 + 1}$$

的最小值.

(2020 BAMO-8: C, BAMO-12: 1 Equation) Find all real numbers $x$ that satisfy the equation $$\frac{x-2020}{1}+\frac{x-2019}{2}+\cdots+\frac{x-2000}{21}=\frac{x-1}{2020}+\frac{x-2}{2019}+\cdots+\frac{x-21}{2000},$$ and simplify your answer(s) as much as possible. Justify your solution.

set A=x-2021, then it would be converted to (A+1)/1+(A+2)/2+(A+3)/3...+(A+21)/21=(A+2020)/2020+(A+2019)/2019+...+(A+2000)/2000. it would
turn out to be A/1+A/2+A/3+....+A/21=A/2020+A/2019+...A/2000. Move everything from right to the left: A/1+A/2+A/3+...+A/21-A/2020-A/2019-...-A/2000=0. then A(1/1+1/2+1/3+...1/21-1/2020-1/2019-....-1/2000)=0. Because (1/1+1/2+1/3+...+1/21-1/2020-1/2019-...-1/2000) not equal 0, A should be 0. thus x=2021

 

一道比较变态的BMT代数题

 

Young不等式

 

一些「经典」不等式的证明（一）