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 \( a+b+c \geq \frac{3}{a+b+c}+\frac{2}{abc} \)
Cezar Lupu si Valentin Vornicu, lista scurta, 2006
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Claudiu Mindrila
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Marius Mainea
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Din conditia initiala \( \sum {a}\ge\sum {\frac{1}{a}}\ge\frac{9}{a+b+c} \) si de aici \( (a+b+c)^2\ge 9 \) sau \( \frac{a+b+c}{3}\ge \frac{3}{a+b+c} \)*
Deasemenea tot din relatia data avem \( (a+b+c)abc\ge ab+bc+ca\ge\sqrt{3abc(a+b+c)} \) sau \( (a+b+c)abc\ge3 \)
adica \( \frac{2(a+b+c)}{3}\ge \frac{2}{abc} \)**
Prin adunarea relatiilor (*) si (**) obtinem concluzia.
Deasemenea tot din relatia data avem \( (a+b+c)abc\ge ab+bc+ca\ge\sqrt{3abc(a+b+c)} \) sau \( (a+b+c)abc\ge3 \)
adica \( \frac{2(a+b+c)}{3}\ge \frac{2}{abc} \)**
Prin adunarea relatiilor (*) si (**) obtinem concluzia.