Áp dụng BĐT BSC:

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

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

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

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

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

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

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

Cộng vế với vế:

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

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

\(5x^2+2xy+2y^2-\left(4x^2+4xy+y^2\right)=\left(x-y\right)^2\ge0\\ \Leftrightarrow5x^2+2xy+2y^2\ge4x^2+4xy+y^2=\left(2x+y\right)^2\)

\(\Leftrightarrow P\le\dfrac{1}{2x+y}+\dfrac{1}{2y+z}+\dfrac{1}{2z+x}=\dfrac{1}{9}\left(\dfrac{9}{x+x+y}+\dfrac{9}{y+y+z}+\dfrac{9}{z+z+x}\right)\\ \Leftrightarrow P\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{x}\right)\\ \Leftrightarrow P\le\dfrac{1}{9}\left(\dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}\right)=\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=1\)

Dấu \("="\Leftrightarrow x=y=z=1\)

\(\sqrt{5x^2+2xy+2y^2}=\sqrt{4x^2+2xy+y^2+x^2+y^2}\ge\sqrt{4x^2+2xy+y^2+2xy}=2x+y\)

\(\Rightarrow\dfrac{1}{\sqrt{5x^2+2xy+2y^2}}\le\dfrac{1}{2x+y}=\dfrac{1}{x+x+y}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}\right)=\dfrac{1}{9}\left(\dfrac{2}{x}+\dfrac{1}{y}\right)\)

Tương tự:

\(\dfrac{1}{\sqrt{5y^2+2yz+2z^2}}\le\dfrac{1}{9}\left(\dfrac{2}{y}+\dfrac{1}{z}\right)\) ; \(\dfrac{1}{\sqrt{5z^2+2zx+2x^2}}\le\dfrac{1}{9}\left(\dfrac{2}{z}+\dfrac{1}{x}\right)\)

Cộng vế:

\(P\le\dfrac{1}{9}\left(\dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}\right)=1\)

\(P_{max}=1\) khi \(x=y=z=1\)

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 \(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le1\left(đpcm\right)\)

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

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