Inegalitati

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alex2008
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Inegalitati

Post by alex2008 »

Sa se arate ca :

a)\( \frac{a^6+b^6}{2}\ge 3a^2b^2-4 \)

b)\( a^6+b^6+1\ge 3a^2b^2 \)
. A snake that slithers on the ground can only dream of flying through the air.
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salazar
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Post by salazar »

a) \( a^6+b^6\ge 2(3a^2b^2-4) \Longleftrightarrow a^6+b^6+8\ge 6a^2b^2 \).
\( a^6+b^6+8=\frac{a^6}{2}+\frac{a^6}{2}+\frac{b^6}{2}+\frac{b^6}{2}+\frac{8}{2}+\frac{8}{2}\ge 6\sqrt[6]{\frac {a^{12}\cdot b^{12}\cdot 64}{2^6=64}}=6a^2b^2 \).
b) \( a^6+b^6+1^6\ge 3\sqrt[3]{a^6b^61^6}=3a^2b^2 \).
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