※ 引述《mathblue (Subriemannian)》之銘言：
: 請證明不可能找到實數a,b,c,d,e
: 使得 對所有x為實數
: 不等式 aCosx+bCos2x+cCos3x+dCos4x+eCos5x > 0 恆成立
若aCosx+bCos2x+cCos3x+dCos4x+eCos5x > 0 恆成立
則aCosX + bCos2X + cCos3X + dCos4X + eCos5X > 0 成立---(0)
以 90-X , 90 + X 和 180 + X 分別帶入x (以下度都省略)
可得
aCos90+X + bCos180+2X + cCos270+3X + dCos360+4X + eCos450+5X
= -aSinX - bCos2X + cSin3X + dCos4X - eSin5X > 0 ---(1)
aCos90-X + bCos180-2X + cCos270-3X + dCos360-4X + eCos450-5X
=aSinX - bCos2X - cSin3X + dCos4X + eSin5X > 0 ---(2)
(1) + (2) =>
dCos4X - bCos2X > 0 ---(3)
aCos180+X + bCos360+2X + cCos540+3X + dCos720+4X + eCos900+5X
=aCos180+X + bCos360+2X + cCos540+3X + dCos720+4X + eCos900+5X
=aCos180+X + bCos2X + cCos180+3X + dCos4X + eCos180+5X
=-aCosX + bCos2X - cCos3X + dCos4X - eCos5X > 0 ---(4)
(0) + (4) =>
dCos4X + bCos2X > 0 ---(5)
(3) + (5) =>
dCos4X > 0 -><-
故假設有誤 得證
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