a>b>c>0且a+b=c,求证a^2/3+b^2/3>c^2/3

(a^2/3+b^2/3)^3=a^2+b^2+3a^(4/3)*b^(2/3)+3*a^(2/3)*b^(4/3)
a+b=c
a^2+b^2+2ab=c^2
a^2+b^2=c^2-2ab
所以(a^2/3+b^2/3)^3=c^2-2ab+3a^(4/3)*b^(2/3)+3*a^(2/3)*b^(4/3)

3a^(4/3)*b^(2/3)+3*a^(2/3)*b^(4/3)>=3*2根号[a^(4/3)*b^(2/3)*a^(2/3)*b^(4/3)]=6根号(a^2b^2)=6ab
所以(a^2/3+b^2/3)^3>=c^2-2ab+6ab=c^2+4ab>c^2=[c^(2/3)^3
所以a^2/3+b^2/3>c^2/3

把a/c设为某角的正弦平方，b/c设为该角的余弦平方。巧妙运用1这个数。