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$a^2(1+b^2)+b^2(1+c^2)+(c^2(1+a^2)\geq 6abc$

♂Vitamin_Tờ♫ · Answer 1 · 7/23/2013 10:00:01 PM

$VT=a^2+a^2b^2+b^2+b^2c^2+c^2+c^2a^2\geq 6\sqrt[6]{(abc)^6}=6abc$

Trả lời 23-07-13 10:00 PM

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p3.luv.0ppasuju · Answer 2 · 7/23/2013 11:30:05 PM

) a²(1+b²)+b²(1+c²)+c²(1+a²)³ 6abc

Û a²+a²b²+b²+b²c²+c²+c²a²-6abc³ 0

Û (a-bc)²+(b-ac)²+(c-ab)²³ 0

Baát ñaúng thöùc cuoái ñuùng, suy ra : a²(1+b²)+b²(1+c²)+c²(1+a²)³ 6abc

Trả lời 23-07-13 11:30 PM

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