Fie a,b,c>0 astfel incat \( abcd(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})=a+b+c+d. \) Sa se arate ca :
\( \frac{(a+1)^2}{a+b}+\frac{(b+1)^2}{b+c}+\frac{(c+1)^2}{c+d}+\frac{(d+1)^2}{d+a}\le 2(a+b+c+d). \)
Inegalitate conditionata 3
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Marius Mainea
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Din conditie avem : \( abc + bcd + cda + dab = a + b + c + d \)
Atunci : \( (a + b)(a + c)(a + d) = (a^2 + 1)(a + b + c + d) \) , deci \( \frac {a^2 + 1}{a + b} = \frac {(a + c)(a + d)}{a + b + c + d} \)
Astfel : \( \frac {a^2 + 1}{a + b} + \frac {b^2 + 1}{b + c} + \frac {c^2 + 1}{c + d} + \frac {d^2 + 1}{d + a} = a + b + c + d \)
Acum mai folosim inegalitatea \( a^2 + 1 \ge \frac {(a + 1)^2}{2} \)
Atunci : \( (a + b)(a + c)(a + d) = (a^2 + 1)(a + b + c + d) \) , deci \( \frac {a^2 + 1}{a + b} = \frac {(a + c)(a + d)}{a + b + c + d} \)
Astfel : \( \frac {a^2 + 1}{a + b} + \frac {b^2 + 1}{b + c} + \frac {c^2 + 1}{c + d} + \frac {d^2 + 1}{d + a} = a + b + c + d \)
Acum mai folosim inegalitatea \( a^2 + 1 \ge \frac {(a + 1)^2}{2} \)
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