已知α、β、γ为锐角,且(sin
设 a=sinα,b=sinβ ,c=sinγ,
则a, b, c ∈(0, 1)
a^3+b^3+c^3 =1 ,
tan^2α=sin^2α/(1-sin^2α) =a^2/(1-a^2)
(tanα)^2+(tanβ)^2+(tanγ)^2
=a^2/(1-a^2)+b^2/(1-b^2) +c^2/(1-c^2)
=a^3/(a-a^3) +b^3/(b-b^3)+c^3/(c-c^3)
a-a^3=1/√2*√[2a^2(1-a^2)^2)] ≤1/√2*√((2a^2+1-a^2+1-a^2)/3)^3=2√3/9
同理b-b^3≤2√3/9,c-c^3≤ 2√3/9
所以
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设 a=sinα,b=sinβ ,c=sinγ,
则a, b, c ∈(0, 1)
a^3+b^3+c^3 =1 ,
tan^2α=sin^2α/(1-sin^2α) =a^2/(1-a^2)
(tanα)^2+(tanβ)^2+(tanγ)^2
=a^2/(1-a^2)+b^2/(1-b^2) +c^2/(1-c^2)
=a^3/(a-a^3) +b^3/(b-b^3)+c^3/(c-c^3)
a-a^3=1/√2*√[2a^2(1-a^2)^2)] ≤1/√2*√((2a^2+1-a^2+1-a^2)/3)^3=2√3/9
同理b-b^3≤2√3/9,c-c^3≤ 2√3/9
所以
a^3/(a-a^3)+b^3/(b-b^3)+c^3/(c-c^3)≥3√3/2(a^3+b^3+c^3)=3√3/2
即 (tanα)^2+(tanβ)^2+(tanγ)^2≥(3√3)/2。
。收起