Áp dụng bđt phụ \(\dfrac{1}{A+B}\le\dfrac{1}{4}\left(\dfrac{1}{A}+\dfrac{1}{B}\right)\forall A,B>0\)

\(\dfrac{1}{2x+y+z}=\dfrac{1}{\left(x+y\right)+\left(x+z\right)}\le\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z}\right)\) Tương tự: \(\dfrac{1}{x+2y+z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

\(\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

\(\Rightarrow\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{4}{x}+\dfrac{4}{y}+\dfrac{4}{z}\right)=1\)

Dấu bằng xảy ra \(\Leftrightarrow x=y=z=\dfrac{3}{4}\)

Áp dụng BĐT BSC:

\(F=\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\)

\(\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}\right)\)

\(=\dfrac{1}{16}\left(\dfrac{4}{x}+\dfrac{4}{y}+\dfrac{4}{z}\right)=\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{1}{4}.4=1\)

\(maxF=1\Leftrightarrow x=y=z=\dfrac{3}{4}\)

Ta cần chứng minh: 

\(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\left(1\right)\left(a,b>0\right)\)

\(\Leftrightarrow\dfrac{4}{a+b}\le\dfrac{a+b}{ab}\\ \Leftrightarrow4ab\le\left(a+b\right)^2\\ \Leftrightarrow\left(a-b\right)^2\ge0\left(luôn.đúng\right)\)

\(DBXR\Leftrightarrow a=b\)

Do các phép biến đổi tương đương nên (1) luôn đúng

Áp dụng (1), ta có:

\(\dfrac{1}{2x+y+z}\le\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\le\dfrac{1}{4}\left[\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{z}\right)\right]=\dfrac{1}{16}\left(\dfrac{2}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

Chứng minh tương tự, ta được:

\(\dfrac{1}{x+2y+z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}\right)\)

\(\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{z}\right)\)

Cộng từng vế BĐT, ta được:

\(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le\dfrac{1}{16}.\left(\dfrac{4}{x}+\dfrac{4}{y}+\dfrac{4}{z}\right)=\dfrac{1}{4}.\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{1}{4}.4=1\)Hay \(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le1\left(đpcm\right)\)

\(DBXR\Leftrightarrow x=y=z=\dfrac{3}{4}\)

Bổ đề:\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\Leftrightarrow\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)

Ta có:\(\dfrac{1}{2x+y+z}\le\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\le\dfrac{1}{4}.\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z}\right)\)

Tương tự ta có:\(\dfrac{1}{2y+z+x}\le\dfrac{1}{4}.\dfrac{1}{4}\left(\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{x}\right)\)

                         \(\dfrac{1}{2z+x+y}\le\dfrac{1}{4}.\dfrac{1}{4}\left(\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{z}+\dfrac{1}{y}\right)\)

Cộng vế với vế ta có:

\(\dfrac{1}{2x+y+z}+\dfrac{1}{2y+z+x}+\dfrac{1}{2z+x+y}\le\dfrac{1}{16}\left[4\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\right]=\dfrac{1}{16}.4.4=1\)

Dấu "=" xảy ra ⇔ \(x=y=z=\dfrac{3}{4}\)

Áp dụng BĐT : \(\dfrac{1}{x}+\dfrac{1}{y}\) \(\geq \) \(\dfrac{4}{x+y}\) \(\Leftrightarrow\) \(\dfrac{1}{4}.\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\) \(\geq\) \(\dfrac{1}{x+y}\)

Ta có: \(\dfrac{1}{2x+y+z}\)=\(\dfrac{1}{\left(x+y\right)+\left(x+z\right)}\)\(\leq\)\(\dfrac{1}{4}.\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\)\(\leq\)\(\dfrac{1}{4}\left(\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+\dfrac{1}{4}\left(\dfrac{1}{x+z}\right)\right)\)=\(\dfrac{1}{16}\left(\dfrac{2}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)(1)

Chứng minh tương tự,ta có:

\(\dfrac{1}{x+2y+z}\) \(\leq\) \(\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}\right)\)(2)

\(\dfrac{1}{x+y+2z}\) \(\leq\) \(\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{z}\right)\)(3)

Đặt: \(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\) là VT

Cộng các BĐT(1),(2),(3) lại với nhau ta được:

VT \(\leq\)\(\dfrac{1}{16}\left(\dfrac{2}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}\right)+\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{z}\right)\)

\(\Leftrightarrow\) VT \(\leq\) \(\dfrac{1}{16}\left(\dfrac{4}{x}+\dfrac{4}{y}+\dfrac{4}{z}\right)\)=\(\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)=\(\dfrac{1}{4}.4=1\)

\(\Leftrightarrow\) \(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\) \(\leq\) 1

Dấu = xảy ra khi x=y=z=\(\dfrac{3}{4}\)

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Câu hỏi của Dương Thị Thu Ngọc - Toán lớp 9 | Học trực tuyến

Ta có: \(\dfrac{x^3}{y+2z}+\dfrac{y^3}{z+2x}+\dfrac{z^3}{x+2y}=\dfrac{x^4}{xy+2zx}+\dfrac{y^4}{yz+2xy}+\dfrac{z^4}{zx+2yz}\)

\(\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{xy+2zx+yz+2xy+zx+2yz}=\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\)

Mà ta lại có: \(xy+yz+zx\le x^2+y^2+z^2\)

 \(\Rightarrow\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1^2}{3.1}=\dfrac{1}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{1}{\sqrt{3}}\)

Thay $x=\sqrt{\frac{1}{2,5}}; y=z=\sqrt{\frac{1}{0,25}}$ ta thấy đề sai bạn nhé!