三角不等式

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t为锐角,求证1+1/sint)*(1+1/cost)>5.

证明 上述待证不等式右边系数不是最佳的,最佳系数为(1+√2)^2，当t=π/4取得.即

(1+1/sint)*(1+1/cost)≥(1+√2)^2.  (1)

(1)左边=1+(sint+cost)/(sint*cost)+1/(sint*cost)

≥1+2/√(sint*cost)+1/(sint*cost)

=[1+1/√(sint*cost)]^2=[1+(√2)/√(sin2t)]^2≥(1+√2)^2.

显然(1+√2)^2>5

证明:由柯西不等式,得

(1+1/sint)*(1+1/cost)≥(1+1/√(sintcost))2

≥(1+1/√((sin2t+cos2t)/2))2

=(1+√2)2

>5

(1+1/sint)(1+1/cost)=1+1/sint+1/cost+1/sint*cost

因为t为锐角

所以sint,cost>0

注:下面式子里sqrt()的意思为开根号

原式>=1+2*sqrt(1/sint*cost)+1/sint*cost>=1+2*sqrt(2)+2=1+2*1.414+2>5

原不等式等价于(1+sint)(1+cost)>5sintcost

即1+sint+cost>4sintcost

令sint+cost=x,(12x^2-2即2x^2-x-3

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