Question

cho x,y,z >0 thoả mãn \(\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)=8\)

tìm Max: \(P=\dfrac{x^2+y^2+z^2+14xyz}{4\left(x+y+z\right)+15xyz}\)

Answer 1

\(\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)=8\)

=>\(8xyz=xyz+\sum x+\sum xy+1\)

=>\(\sum x^2+14xyz=\left(\sum x\right)^2+2\sum x+2\)

mặt khác

\(8=\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)\ge\dfrac{8}{\sqrt[3]{xyz}}\rightarrow xyz\ge1\)

đặt \(\sum x=a\left(a\ge3\right)\)

khi đó \(P=\dfrac{a^2+2a+2}{4a^2+15xyz}\le\dfrac{a^2+2a+2}{4a^2+15}\)

\(\dfrac{a^2+2a+2}{4a^2+15}=\dfrac{1}{3}-\dfrac{\left(a-3\right)^2}{12a^2+45}\le\dfrac{1}{3}\)

vậy max bằng 1/3 khi x=y=z=1

Answer 2

@Lightning Farron @Akai Haruma @Vũ Tiền Châu

Answer 3

xin lỗi mọi người nhá. dướng mẫu là 4(x+y+z)^2