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证明:对于任意的a,b,c,d∈R,恒有不等式(ac+bd)~2≤(a~2+b~2)(c~2+d~2).(必修4A版P108B组第3题)此处的证明思路是构造两个平面向量α=(a,b),β=(c,d),利用不等式|α·β|≤|α|·|β|即得,等号当且仅当向量α,β共线,即ad=bc时成立,于是有(ac+bd)~2=(a~2+b~2)(c~2+d~2).上述不等式其实就是二维形式的柯西不等式,它在求几类三角函数的最值上有着很重要的作用,笔者将其整理出来,与大家共赏.1.求函数y=asinx+bcosx(ab≠0)的最值这类函数的最值用公式y=asinx+bcosx
It is proved that the constant inequality (ac + bd) ~ 2 ≤ (a ~ 2 + b ~ 2) (c ~ 2 + d ~ 2) for any a, b, c, d∈R. Problem 3) The proof here is to construct two plane vectors α = (a, b), β = (c, d), using the inequality | α · β | ≤ | α | · | β | (Ac + bd) ~ 2 = (a ~ 2 + b ~ 2) (c ~ 2 + d ~ 2) if and only if the vectors α, β are collinear, that is, ad = bc. Inequality is actually two-dimensional form of Cauchy inequality, it is in seeking the most value of several types of trigonometric functions have a very important role, I will put it out, and we share.1. Find the function y = asinx + bcosx (ab ≠ 0) The most value of such functions using the formula y = asinx + bcosx