Question

如图,△ABC的三边分别为abc,BC边上高为h,求证b+c≤根号a²+4h²

并说明在什么情况下等号成立、？
答对追分

Answer 1

令a=a1±a2;a1,h,c和a2,h,b分别构成直角三角形.则a1^2=c^2-h^2;a2^2=b^2-h^2.
则a^2=(a1±a2)^2
=a1^2±2·a1·a2+a2^2
=b^2+c^2±2·a1·a2-2h^2
=(b+c)^2±2·a1·a2-2(bc+h^2).
则:
(b+c)²=a²+2h²+2[bc-(±a1·a2)]
=a²+2h²+2[bc-(±√(c²-h²)(b²-h²) )]
=a²+2h²+2[bc-(±√(b²·c²-h²·(b²+c²)+h^4) ]
=a²+2h²+2[bc-(±√(b²·c²-h²·(b²+c²)+h^4) ]·[bc+(±√(b²·c²-h²·(b²+c²)+h^4) ]/[bc±√(b²·c²-h²·(b²+c²)+h^4) ]
=a²+2h²+2[b²·c²-(b²·c²-h²·(b²+c²)+h^4)]/[bc±√(b²·c²-h²·(b²+c²)+h^4) ]
=a²+2h²+2h²·{[(b²+c²)-h²]/[bc±√(b²·c²-h²·(b²+c²)+h^4) ]}
=a²+2h²+2h²·{[(b/h)²+(c/h)²-1]/[(b/h)(c/h)±√( (b/h)²+(c/h)²)+1]}
∵{[(b/h)²+(c/h)²-1]/[(b/h)(c/h)±√( (b/h)²+(c/h)²)+1]}≤1,
∴有b+c≤√(a²+4h²)

当[(b/h)²+(c/h)²-1]/[(b/h)(c/h)±√( (b/h)²+(c/h)²)+1]=1,时
即bc=h²时,等号成立.

Answer 2

令a=a1±a2;a1,h,c和a2,h,b分别构成
直角三角形
.则a1^2=c^2-h^2;a2^2=b^2-h^2.
则a^2=(a1±a2)^2
=a1^2±2·a1·a2+a2^2
=b^2+c^2±2·a1·a2-2h^2
=(b+c)^2±2·a1·a2-2(bc+h^2).
则:
(b+c)²=a²+2h²+2[bc-(±a1·a2)]
=a²+2h²+2[bc-(±√(c²-h²)(b²-h²)
)]
=a²+2h²+2[bc-(±√(b²·c²-h²·(b²+c²)+h^4)
]
=a²+2h²+2[bc-(±√(b²·c²-h²·(b²+c²)+h^4)
]·[bc+(±√(b²·c²-h²·(b²+c²)+h^4)
]/[bc±√(b²·c²-h²·(b²+c²)+h^4)
]
=a²+2h²+2[b²·c²-(b²·c²-h²·(b²+c²)+h^4)]/[bc±√(b²·c²-h²·(b²+c²)+h^4)
]
=a²+2h²+2h²·{[(b²+c²)-h²]/[bc±√(b²·c²-h²·(b²+c²)+h^4)
]}
=a²+2h²+2h²·{[(b/h)²+(c/h)²-1]/[(b/h)(c/h)±√(
(b/h)²+(c/h)²)+1]}
∵{[(b/h)²+(c/h)²-1]/[(b/h)(c/h)±√(
(b/h)²+(c/h)²)+1]}≤1,
∴有b+c≤√(a²+4h²)
当[(b/h)²+(c/h)²-1]/[(b/h)(c/h)±√(
(b/h)²+(c/h)²)+1]=1,时
即bc=h²时,等号成立.