(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 $bge 0$, it suffices to prove$$sum_{cyc} dfrac{ab}{3} le dfrac{4}{3} iff (a+b+c+d)^2 ge 4(ab+bc+cd+da)$$$$iff (a-b+c-d)^2ge 0$$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} ight)^2 = 4. ]Thus[ sum_{ ext{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