Cho biểu thức P = cosA+cosB+cosC. Trong đó A,B,C là các góc của tam giác ABC bất kì. Chứng minh rằng P đạt giá trị lớn nhất nhưng không đạt giá trị nhỏ nhất.
P= cosA+cosB+cosC=2cosA+B2cosAB2+12sin2C2
    =1+2sinC2cosAB2+12sin2C2
    =1+2sinC2(cosAB2cosA+B2)=1+4sinC2sinA2sinB2
Từ đó: P=12(sin2C2sinC2cosAB2)
                =12(sinC212cosAB2)2+12cos2AB2
                =322(sinC212cosAB2)212sin2AB232
P=32{sinAB2=0sinC212cosAB2=0{A=BsinC2=12ΔABCđều
Đáp số: max = \frac{3}{2} xảy ra khi và chỉ khi \Delta ABCđều

Mặt khác P = 1 + 4\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2} > 1,\forall \Delta ABC, do đó nếu P đạt giá trị nhỏ nhất bằng m thì ta cũng có P \ge m > 1,\forall \Delta ABC
Xét các tam giác ABC với A = 2x,B = C = \frac{\pi }{2} - x thì P = 1 + 4\sin .{\sin ^2}\left( {\frac{\pi }{4} - \frac{x}{2}} \right);\mathop {\lim P}\limits_{x \to 0}  = \mathop {\lim }\limits_{x \to 0} \left( {1 + 4\sin x.{{\sin }^2}\left( {\frac{\pi }{4} - \frac{x}{2}} \right)} \right) = 1 + 0 = 1
Nhưng với P \ge m,\forall \Delta ABC \Rightarrow \mathop {\lim P}\limits_{x \to 0}  \ge m \Rightarrow 1 \ge m (vô lý) .
Vậy P không có giá trị nhỏ nhất
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