[ Toán Khó ] Tìm Giá Trị Lớn Nhất !

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tcbn01

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quanghao98


Đặt a=x3;b=y3;c=z3;x,y,z>0a=x^3;b=y^3;c=z^3;x,y,z>0xyz=1xyz=1
Ta sẽ chứng minh BDT sau:
x3y6+z6+x3+y3x6+z6+y3+z3x6+y6+z3\dfrac{x^3}{y^6+z^6+x^3}+\dfrac{y^3}{x^6+z^6+y^3}+\dfrac{z^3}{x^6+y^6+z^3}\leq 11
(y5z5)(yz)(y^5-z^5)(y-z)\geq 00

\Leftrightarrow y6+z6y^6+z^6\geqy5z+z5yy^5z+z^5y
\Rightarrow y6+z6+x4yzy^6+z^6+x^4yz\geqyz(x4+y4+z4)yz(x^4+y^4+z^4)

\Rightarrow y6+z6+x3y^6+z^6+x^3\geq yz(x4+y4+z4)yz(x^4+y^4+z^4)

\Rightarrow 1y6+z6+x3\dfrac{1}{y^6+z^6+x^3}\leq 1yz(x4+y4+z4)\dfrac{1}{yz(x^4+y^4+z^4)}

\Rightarrow x4yzy6+z6+x3\dfrac{x^4yz}{y^6+z^6+x^3}\leqx4x4+y4+z4\dfrac{x^4}{x^4+y^4+z^4}

\Rightarrow x3y6+z6+x3\dfrac{x^3}{y^6+z^6+x^3}\leq x4x4+y4+z4\dfrac{x^4}{x^4+y^4+z^4}
Thiết lập các BDT tương tự ta có DPCM,MAx=1MAx=1 khi a=b=c=1a=b=c=1
 
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