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\[2\sqrt{log_{19}18\times log_{19}20}\le log_{19}18+log_{19}20=log_{19}360<2 \]
所以$log_{19}18\times log_{19}20< 1$即x>y
z>$\frac{19}{18}>x$
即要证\[18^{\frac{1}{18}}>19^{\frac{1}{19}}\]
即\[\frac{1}{18}\ln 18>\frac{1}{19}\ln 19\]
构造函数$y=\frac{\ln x}{x}$,求导,在x>e单调递减 |
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