a_507_bc wrote: If $a, b, c, d>0$ and $abcd=1$, show that $$\frac{ab+1}{a+1}+\frac{bc+1}{b+1}+\frac{cd+1}{c+1}+\frac{da+1}{d+1} \geq 4. $$When does equality hold? https://artofproblemsolving.com/community/c6h551158p3199527 https://artofproblemsolving.com/community/c6h572877p3370410 anhduy98 wrote: valsidalv007 wrote: Let $a$,$b$,$c$,$d$ be positive real numbers such that $abcd$=1. Prove that $$\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+cd}{1+c}+\frac{1+da}{1+d} \geq 4$$ $$LHS=\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+\dfrac{1}{ab}}{1+c}+\frac{1+\dfrac{1}{bc}}{1+\dfrac{1}{abc}}$$$$LHS=\frac{1+ab}{1+a}+\frac{1+ab}{ab+abc}+\frac{a(1+bc)}{a+ab}+\frac{a(1+bc)}{1+abc}$$$$LHS \ge \frac{4(1+ab)}{1+a+ab+abc}+\frac{4a(1+bc)}{1+a+ab+abc}=4=RHS$$ https://artofproblemsolving.com/community/c6h2010928p14101614 Team Competition Aug. 24, 2023 Individual Competition Aug. 23, 2023

Identical to a problem from Bulgaria RMM TST 2016. We may write $a=\frac{y}{x}$, $b=\frac{z}{y}$, $c=\frac{t}{z}$ and $d=\frac{x}{t}$ for $x, y, z, t > 0$. Then $\sum_{cyc}\frac{1+ab}{1+a}=\sum_{cyc}\frac{x+z}{x+y} =(x+z)\left(\frac{1}{x+y}+\frac{1}{z+t}\right)+(y+t)\left(\frac{1}{y+z}+\frac{1}{t+x}\right)\geq \frac{4(x+z)+4(y+t)}{x+y+z+t}=4$, done.

a_507_bc wrote: If $a, b, c, d>0$ and $abcd=1$, show that $$\frac{ab+1}{a+1}+\frac{bc+1}{b+1}+\frac{cd+1}{c+1}+\frac{da+1}{d+1} \geq 4. $$When does equality hold? Assassino9931 wrote: Identical to a problem from Bulgaria RMM TST 2016. We may write $a=\frac{y}{x}$, $b=\frac{z}{y}$, $c=\frac{t}{z}$ and $d=\frac{x}{t}$ for $x, y, z, t > 0$. Then $\sum_{cyc}\frac{1+ab}{1+a}=\sum_{cyc}\frac{x+z}{x+y} =(x+z)\left(\frac{1}{x+y}+\frac{1}{z+t}\right)+(y+t)\left(\frac{1}{y+z}+\frac{1}{t+x}\right)\geq \frac{4(x+z)+4(y+t)}{x+y+z+t}=4$, done. I'd like to add that equality holds if and only if $x+y=z+t,y+z=t+x$, i.e. $x=z,y=t$, i.e. $a=c=\frac{1}{b}=\frac{1}{d}$.