Inegalitatea 1

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Cezar Lupu
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Inegalitatea 1

Post by Cezar Lupu »

Fie \( a, b, c \) trei numere reale strict pozitive astfel incat \( a+b+c\geq\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \). Sa se arate ca
\( ab+bc+ca\geq 3 \).
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
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Filip Chindea
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Post by Filip Chindea »

\( ab + bc + ca \le abc(a + b + c) = \frac{1}{3}\cdot 3((ab)(bc) + (bc)(ca) + (ca)(ab)) \le \)
\( \frac{1}{3} \cdot (ab + bc + ca)^2 \), de unde concluzia e imediata.
Life is complex: it has real and imaginary components.
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