设$a,b,c>0,$满足$a+b+cle abc$证明:$dfrac{1}{sqrt{1+a^2}}+dfrac{1}{sqrt{1+b^2}}+dfrac{1}{sqrt{1+c^2}}ledfrac{3}{2}$
证明:设$a=dfrac{1}{x},b=dfrac{1}{y},c=dfrac{1}{z}$由$a+b+cle abc$知$xy+yz+zxle 1$
egin{align}label{}
sumdfrac{1}{sqrt{1+a^2}}&=sumdfrac{x}{sqrt{1+x^2}} \
&lesumdfrac{x}{sqrt{x^2+xy+yz+zx}}\
&=sumdfrac{x}{sqrt{(x+y)(x+z)}}\
&=sumdfrac{xsqrt{(x+y)(x+z)}}{(x+y)(x+z)}\
&lesumdfrac{x(x+y+x+z)}{2(x+y)(x+z)}\
&=sumdfrac{1}{2}left(dfrac{x}{x+y}+dfrac{x}{x+z}
ight)\
&=dfrac{3}{2}\
end{align}.
注:
如果条件改为$a+b+c=abc，a>0,b>0,c>0$则容易想到$x=tanA,y=tanB,z=tanC$的代换.
其中$A,B,C$为锐角三角形的三个角.
另一个常见的代换$p,q,rge0,p^2+q^2+r^2+2pqr=1$时，
可令$p=cosA,q=cosB,r=cosC,A,B,Cin[0,dfrac{pi}{2}],A+B+C=pi$