a=√（x^2-xy+y^2）,b=p√（xy）,c=x+y,若对任意的正实数x,y,都存在以a、b、c为三边长的三角形,求p范围

Question

a=√（x^2-xy+y^2）,b=p√（xy）,c=x+y,若对任意的正实数x,y,都存在以a、b、c为三边长的三角形,求p范围

Answer 1

根据三角形的两边之和大于第三遍,b+c＞a,p√（xy）+x+y＞√（x^2-xy+y^2）,x²+y²+2xy+p²xy+2p(x+y)√（xy）＞x²-xy+y²,3xy+p²xy+2p(x+y)√（xy）＞0,恒成立；a+b＞c,√(x²-xy+y²)+p√(xy)＞x+y,x²-xy+y²+2p√[(x²-xy+y²)xy]+p²xy＞x²+y²+2xy,2p√[(x²-xy+y²)＞(3-p²)√(xy),x²-xy+y²＞(3-p²)²xy/4p²,x²+y²＞[(3-p²)²+4p²]xy/4p²,当[(3-p²)²+4p²]/4p²＜2时,上式恒成立,解得：1＜p＜3；a+c＞b,√(x²-xy+y²)+x+y＞p√(xy),2x²+2y²+xy+2(x+y)√(x²-xy+y²)＞p²xy,2x²+2y²+xy+2(x+y)√(x²-xy+y²)＞4xy+xy+4√[(xy)(2xy-xy)]=9xy,9xy＞p²xy,-3＜p＜3；综上p范围为(1,3).