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Числа $%a,b,c\in (0;1)$% и $%x,y,z\in(0;∞)$% удовлетворяют условиям: $$a^x=bc$$ $$b^y=ac$$ $$c^z=ab$$ Докажите что: $$\frac{1}{2+x} + \frac{1}{2+y} + \frac{1}{2+z}\leq \frac{3}{4}$$ задан 29 Май '19 18:27 
potter |
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$$A = \dfrac {1}{a}\ , \ B= \dfrac {1}{b}\ , \ C= \dfrac {1}{c}\ \ (A, B, C > 1)$$ $$ x= \dfrac {ln (B)+ln (C)}{ln (A)}\ ....$$ $$\Leftrightarrow \dfrac {ln (A)}{ln (A)+ln (ABC)}+\dfrac {\ln (B)}{ln (B)+ln (ABC)}\dfrac {ln (C)}{ln (C)+ln (ABC) }\le$$ $$\le3 \dfrac {ln (ABC)/3}{ln (ABC)/3 +ln (ABC)}=\dfrac {3}{4}$$ отвечен 29 Май '19 20:55 
Sergic Primazon |