Fie \( a,b,c,d \) patru numere nenegative . Sa se demonstreze ca :
\(
(1-a+a^2)(1-b+b^2)(1-c+c^2)(1-d+d^2)\ge \left(\frac{abcd+1}{2}\right)^2 \)
Patru variabile
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Mateescu Constantin
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Pentru \( a=b=c=d \) inegalitatea se reduce la \( 2(1-a-a^2)^2\ge 1+a^4 \)
\( \Longrightarrow 2(1-a-a^2)^2-1-a^4\ge 0 \Longleftrightarrow (1-a)^4\ge 0 \)
Folosind acest rezultat avem
\( 4(1-a+a^2)^2(1-b+b^2)^2\ge (1+a^4)(1+b^4) \)
Din inegalitatea \( (1+a^4)(1+b^4)\ge (1+a^2b^2)^2 \) vom avea
\( 2(1-a+a^2)(1-b+b^2)\ge (1+a^2b^2) \)
Analog obtinem si \( 2(1-c+c^2)(1-d+d^2)\ge (1+c^2d^2) \)
Scriem si inegalitatea \( (1+a^2b^2)(a+c^2d^2)\ge (1+abcd)^2 \)
Inmultind ultimele trei relatii obtinem concluzia.
Egalitatea are loc pentru \( a=b=c=d=1 \).
\( \Longrightarrow 2(1-a-a^2)^2-1-a^4\ge 0 \Longleftrightarrow (1-a)^4\ge 0 \)
Folosind acest rezultat avem
\( 4(1-a+a^2)^2(1-b+b^2)^2\ge (1+a^4)(1+b^4) \)
Din inegalitatea \( (1+a^4)(1+b^4)\ge (1+a^2b^2)^2 \) vom avea
\( 2(1-a+a^2)(1-b+b^2)\ge (1+a^2b^2) \)
Analog obtinem si \( 2(1-c+c^2)(1-d+d^2)\ge (1+c^2d^2) \)
Scriem si inegalitatea \( (1+a^2b^2)(a+c^2d^2)\ge (1+abcd)^2 \)
Inmultind ultimele trei relatii obtinem concluzia.
Egalitatea are loc pentru \( a=b=c=d=1 \).