已知：sin^4x/a+cos^4x/b=1/(a+b) (a>0,b>0)
1个回答
证明 由 sin^4x/a +cos4^x/b =1(a+b),
得( a+b)/a sin^4x +(a+b)/b cos^4x=1,
即 b/asin^4x+a/bcos^4x+sin4x+cos^4x=1.
又 sin^4x +cos^4x =(sin² x +cos² x ) ² -2 sin² xcos² x=1- 2 sin² xcos² x,
则 b/asin^4x+a/bcos^4x- 2 sin ² xcos ² x=0 ,
即 b/asin^4x+a/bcos^4x- 2[根号下 ( b/a)]sin² x [根号下(a/b)]cos² x=0 .则{ [根号下(b/a)]sin² x- [根号下（b/a）]cos² x) }²=0 ,
{[根号下( b/a)]sin² x - [根号下(a/b)]cos² x}² =0,
所以有sin² x = a/(a+,b).cos²x = b/(a+b).
带入不等式 易得.
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