Giả sử x,y,z là các số dương thay đổi tm đk:$ xy^{2}z^{2}+x^{2}z+y=3z^{2}$
Tìm MAX của $ P=\frac{z^{4}}{1+z^{4}(x^{4}+y^{4})}$
bài này cũ rồi thì phải –  ๖ۣۜDevilღ 03-07-17 04:04 PM

Câu hỏi này được treo giải thưởng trị giá +1000 vỏ sò bởi tranquynhat2002@gmail.com, đã hết hạn vào lúc 03-07-17 02:08 AM

$xy^{2}z^{2}+x^{2}z+y=3z^{2}\Leftrightarrow xy^{2}+\frac{x^{2}}{z}+\frac{y}{z^{2}}=3$
$\Leftrightarrow 3\leq (x^{2}y^{2}+y^{2}+\frac{x^{2}}{z^{2}}+x^{2}+\frac{y^{2}}{z^{2}}+\frac{1}{z^{2}})/2$
Đặt $a=x^{2};b=y^{2};c=\frac{1}{z^{2}}(x,y,z>0)\Rightarrow a+b+c+ab+bc+ca\geq 6$
Ta có
$P=\frac{z^{4}}{1+z^{4}(x^{4}+y^{4})}=\frac{1}{\frac{1}{z^{4}}+x^{4}+y^{4}}=\frac{1}{a^{2}+b^{2}+c^{2}}$
Lại có
$a^{2}+1\geq 2a\Rightarrow a^{2}+b^{2}+c^{2}+1+1+1\geq 2a+2b+2c$
$a^{2}+b^{2}+b^{2}+c^{2}+a^{2}+c^{2}\geq 2ab+2bc+2ca$
$\Rightarrow 3(a^{2}+b^{2}+c^{2})+3\geq 2(a+b+c+ab+bc+ca)\geq 12\Rightarrow a^{2}+b^{2}+c^{2}\geq 3$
$\Rightarrow P\leq 1/3$
MaxP=1/3
Dấu bằng xẩy ra khi và chỉ khi x=y=z=1

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