Áp dụng Holder ... Đặt $ VT=A $ ta có:
$ A^2 ( \sum a^2(b+c)^2) \geq (a^2+b^2+c^2)^3 $
Ta sẽ CM $ (a^2+b^2+c^2)^3 \geq \dfrac{9}{4}\sqrt{\dfrac{a^4+b^4+c^4}{3}}(\sum a^2(b+c)^2) $
bdt $ <=> \dfrac{a^2+b^2+c^2}{\sqrt{3(a^4+b^4+c^4)}} \geq \dfrac{3(\sum a^2(b+c)^2)}{4(a^2+b^2+c^2)^2} $
$ 3(\sum a^2(b+c)^2) - 4(a^2+b^2+c^2)^2 = -\sum (b-c)^2(3a^2+2(b+c)^2) $
bdt
$ <=> \sum (b-c)^2 ( \dfrac{3a^2+2(b+c)^2}{4(a^2+b^2+c^2)^2} - \dfrac{(b+c)^2}{\sqrt{3(a^4+b^4+c^4)}(\sqrt{3(a^4+b^4+c^4)}+a^2+b^2+c^2)} \geq 0 $
Mà $ \sqrt{3(a^4+b^4+c^4)}(\sqrt{3(a^4+b^4+c^4)+a^2+b^2+c^2}) \geq 2(a^2+b^2+c^2)^2
=> ( \dfrac{3a^2+2(b+c)^2}{4(a^2+b^2+c^2)^2} - \dfrac{(b+c)^2}{\sqrt{3(a^4+b^4+c^4)}(\sqrt{3(a^4+b^4+c^4)}+a^2+b^2+c^2)} \geq \dfrac{3a^2+(b+c)^2}{4(a^2+b^2+c^2)^2} - \dfrac{(b+c)^2}{2(a^2+b^2+c^2)^2} = \dfrac{ 3a^2 }{4(a^2+b^2+c^2)^2} \geq 0 $ =>
$ A^2 \geq \dfrac{9}{4}\sqrt{\dfrac{a^4+b^4+c^4}{3}} <=> A \geq \dfrac{3}{2} \sqrt[4]{\dfrac{a^4+b^4+c^4}{3}} $
Vậy ta có dpcm ^^ !!
Bài viết đã được chỉnh sửa nội dung bởi MyLoveIs4Ever: 23-04-2008 - 10:50