Inegalitate exponentiala conditionata

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Andrei Velicu
Euclid
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Joined: Wed Oct 17, 2007 9:20 am
Location: Constanta

Inegalitate exponentiala conditionata

Post by Andrei Velicu »

Daca \( a, b>0 \) si \( a+b=1 \), aratati ca \( a^a\cdot b^b\geq \frac{1}{2} \).

Marius Cavachi
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Marius Mainea
Gauss
Posts: 1077
Joined: Mon May 26, 2008 2:12 pm
Location: Gaesti (Dambovita)

Post by Marius Mainea »

Indicatie solutie ,,sofisticata'':

Se deconditioneaza inegalitatea (\( a=\frac{x}{x+y},b=\frac{y}{x+y}) \) si apoi se foloseste faptul ca f(x)=xlnx este convexa.
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