设△ABC为任意三角形求证tan?A/2+tan?B/2+tan?C/2≥1

2tan?A/2+2tan?B/2+2tan?C/2-2tanA/2tanB/2-2tanB/2tanC/2-2tanA/2tanC/2 =(tan?A/2-2tanA/2tanB/2+tan?B/2)+(tan?B/2-2tanB/2tanC/2+tan?C/2)+(tan?A/2-2tanA/2tanC/2+tan?C/2) =(tanA/2-tanB/2)?+(tanB/2-tanC/2)?+(tanA/2-tanC/2)?>=0∴tan?A/2+tan?B/2+tan?C/2-tanA/2tanB/2-tanB/2tanC/2-tanA/2tanC/2>=0∴tan?A/2+tan?B/2+tan?C/2>=tanA/2tanB/2+tanB/2tanC/2+tanA/2tanC/2∵cotA/2=tan(90°-A/2)=tan(180°-A)/2=tan(B+C)/2=tan(B/2+C/2)=(tanB/2+tanC/2)/(1-tanB/2tanC/2)又∵cotA/2tanA/2=1∴tanA/2(tanB/2+tanC/2)/(1-tanB/2tanC/2)=1∴tanA/2tanB/2+tanB/2tanC/2+tanA/2tanC/2=1∴tan?A/2+tan?B/2+tan?C/2>=tanA/2tanB/2+tanB/2tanC/2+tanA/2tanC/2=1 即tan?A/2+tan?B/2+tan?C/2≥1

tan??是啥意思？