Question

[Tuyển Trần Thị]

cho x,y,z là các số thực dương tm xyz=8

cmr \(\dfrac{1}{2x+y+6}+\dfrac{1}{2y+z+6}+\dfrac{1}{2z+x+6}\le\dfrac{1}{4}\)

Answer 1

[Akai Haruma]

Lời giải:

\(\frac{1}{2x+y+6}+\frac{1}{2y+z+6}+\frac{1}{2z+x+6}\leq \frac{1}{4}\)

\(\Leftrightarrow \frac{6}{2x+y+6}+\frac{6}{2y+z+6}+\frac{6}{2z+x+6}\leq \frac{3}{2}\)

\(\Leftrightarrow 1-\frac{2x+y}{2x+y+6}+1-\frac{2y+z}{2y+z+6}+1-\frac{2z+x}{2z+x+6}\leq \frac{1}{4}\)

\(\Leftrightarrow A=\frac{2x+y}{2x+y+6}+\frac{2y+z}{2y+z+6}+\frac{2z+x}{2z+x+6}\geq \frac{3}{2}\)

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Thật vậy. Áp dụng BĐT Cauchy-Schwarz:

\(A=\frac{(2x+y)^2}{(2x+y)(2x+y+6)}+\frac{(2y+z)^2}{(2y+z)(2y+z+6)}+\frac{(2z+x)^2}{(2z+x)(2z+x+6)}\)

\(\geq \frac{(2x+y+2y+z+2z+x)^2}{ (2x+y)(2x+y+6)+(2y+z)(2y+z+6)+(2z+x)(2z+x+6)}\)

\(\Leftrightarrow A\geq \frac{9(x+y+z)^2}{5(x^2+y^2+z^2)+4(xy+yz+xz)+18(x+y+z)}\)

Ta sẽ cm \( \frac{9(x+y+z)^2}{5(x^2+y^2+z^2)+4(xy+yz+xz)+18(x+y+z)}\geq \frac{3}{2}\)

\(\Leftrightarrow \frac{3(x+y+z)^2}{5(x^2+y^2+z^2)+4(xy+yz+xz)+18(x+y+z)}\geq \frac{1}{2}\)

\(\Leftrightarrow x^2+y^2+z^2+8(xy+yz+xz)\geq 18(x+y+z)\)

\(\Leftrightarrow (x+y+z)^2+6(xy+yz+xz)\geq 18(x+y+z)(*)\)

Theo BĐT AM-GM: \((xy+yz+xz)^2\geq 3xyz(x+y+z)\)

\(\Leftrightarrow (xy+yz+xz)^2\geq 24xyz\Rightarrow xy+yz+xz\geq 2\sqrt{6(x+y+z)}\)

Đặt \(\sqrt{6(x+y+z)}=t\)

Có \((x+y+z)^2+6(xy+yz+xz)\geq \frac{t^4}{36}+12t\geq 18.\frac{t^2}{6}\)

\(\Leftrightarrow \frac{t^3}{36}+12\geq 3t\)

\(\Leftrightarrow t^3-108t+432\geq 0\)

\(\Leftrightarrow (t-6)^2(t+12)\geq 0\) (luôn đúng với mọi \(t\geq 0\) )

Do đó ta có \((*)\), từ \((*)\Rightarrow A\geq \frac{3}{2}\). CM kết thúc

Dấu bằng xảy ra khi \(x=y=z=2\)

Answer 2

[Tuyển Trần Thị]

or cách khác

https://olm.vn/hoi-dap/question/1148838.html