Inegalitatea 1, conditionata, cu abc=1 (Crux)
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Cezar Lupu
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Inegalitatea 1, conditionata, cu abc=1 (Crux)
Fie \( a, b, c \) trei numere trict pozitive astfel incat \( abc=1 \). Sa se arate ca \( \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3\geq a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \).
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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Sabin Salajan
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Deconditionam: \( a=\frac{x}{y} \);\( b=\frac{z}{x} \);\( c=\frac{y}{z} \)
Inegalitatea devine echivalenta cu: \( \sum \frac{x^2}{yz}+3\geq\sum \frac{x}{y}+\sum\frac{y}{x} \)
Prin aducere la acelasi numitor (anume xyz, care se simplifica fiind pozitiv) devine:\( \sum x^3+3xyz\geq\sum{x^2}(y+z) \)
care este inegalitatea lui Schur pt n=1 (sau se reduce imediat la cunoscuta \( (x-y+z)(x+y-z)(-x+y+z)\leq xyz \)).
Inegalitatea devine echivalenta cu: \( \sum \frac{x^2}{yz}+3\geq\sum \frac{x}{y}+\sum\frac{y}{x} \)
Prin aducere la acelasi numitor (anume xyz, care se simplifica fiind pozitiv) devine:\( \sum x^3+3xyz\geq\sum{x^2}(y+z) \)
care este inegalitatea lui Schur pt n=1 (sau se reduce imediat la cunoscuta \( (x-y+z)(x+y-z)(-x+y+z)\leq xyz \)).