Đặt \(\left(x;y;z\right)=\left(2a^2;2b^2;2c^2\right)\Rightarrow abc=1\)

\(VT=\frac{1}{4a^2+2b^2+6}+\frac{1}{4b^2+2c^2+6}+\frac{1}{4c^2+2a^2+6}\)

\(VT=\frac{1}{\left(2a^2+2\right)+\left(2a^2+2b^2\right)+4}+\frac{1}{\left(2b^2+2\right)+\left(2b^2+2c^2\right)+4}+\frac{1}{\left(2c^2+2\right)+\left(2c^2+2a^2\right)+4}\)

\(VT\le\frac{1}{4a+4ab+4}+\frac{1}{4b+4bc+4}+\frac{1}{4c+4ca+4}=\frac{1}{4}\)

Dấu "=" xảy ra khi \(a=b=c=1\) hay \(x=y=z=2\)

Liên tục áp dụng bất đẳng thức \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\) và ta có:

\(\frac{1}{3x+3y+2x}=\frac{1}{2\left(x+y\right)+\left(x+y+2z\right)}\le\frac{1}{4}\left(\frac{1}{2\left(x+y\right)}+\frac{1}{\left(x+z\right)+\left(y+z\right)}\right)\le\frac{1}{8\left(x+y\right)}+\frac{1}{16}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\)

Chứng minh tương tự tạ có:

\(\frac{1}{3x+2y+3z}\le\frac{1}{8\left(z+x\right)}+\frac{1}{16}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\)

\(\frac{1}{2x+3y+3z}\le\frac{1}{8\left(y+z\right)}+\frac{1}{16}\left(\frac{1}{z+x}+\frac{1}{x+y}\right)\)

Suy ra \(VT\le\frac{1}{8}\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)+\frac{1}{8}\left(\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{z+x}\right)=\frac{3}{2}\)

Dấu "=" xảy ra <=> \(x=y=z=\frac{1}{4}\)

\(\frac{1}{x^4}+\frac{1}{y^4}=\frac{x^2}{x^6}+\frac{1}{y^4}\ge\frac{\left(x+1\right)^2}{x^6+y^4}\ge\frac{4x}{x^6+y^4}\)

tương tự

\(\frac{1}{y^4}+\frac{1}{z^4}\ge\frac{4y}{y^6+z^4}\);

\(\frac{1}{z^4}+\frac{1}{x^4}\ge\frac{4z}{z^6+x^4}\);

cộng vế với vế => đpcm

Dấu "=" xảy ra <=> x=y=z=1

Cô si

Cách khác:

\(x^6+y^4\ge2\sqrt{x^6y^4}=2x^3y^2\)

\(\Rightarrow\frac{2}{x^6+y^4}\le\frac{1}{x^2y^2}\)

CMTT , ta có VT \(\le\frac{1}{x^2y^2}+\frac{1}{y^2z^2}+\frac{1}{z^2x^2}\)

Bổ đề: \(a^2+b^2+c^2\ge ab+bc+ca\) ( luôn đúng)

\(\Rightarrow\frac{1}{x^2y^2}+\frac{1}{y^2z^2}+\frac{1}{z^2x^2}\le\frac{1}{x^4}+\frac{1}{y^4}+\frac{1}{z^4}\)

ĐPCM

Dấu " =" xảy ra khi x=y=z=1

Lời giải:

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

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

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

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

-----------------------

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 \(x=y=z=2\)

Áp dụng BĐT AM-GM,ta có:

\(\hept{\begin{cases}x^2+y^2\ge2xy\\y^2+1\ge2y\end{cases}}\)

\(\Rightarrow\frac{1}{x^2+2y^2+3}\le\frac{1}{2xy+2y+2}\)

Chứng minh tương tự,ta có:

\(\frac{1}{y^2+2z^2+3}\le\frac{1}{2yz+2z+2}\)

\(\frac{1}{z^2+2x^2+3}\le\frac{1}{2zx+2x+2}\)

Cộng vế theo vế của các bất đẳng thức,ta có được:

\(VT\le\frac{1}{2}\left(\frac{1}{xy+y+1}+\frac{1}{yz+z+1}+\frac{1}{zx+x+1}\right)\)

Mặt khác,ta lại có được:

\(\frac{1}{xy+y+1}+\frac{1}{yz+z+1}+\frac{1}{zx+x+1}\)

\(=\frac{1}{xy+y+1}+\frac{xy}{xy+y+1}+\frac{y}{xy+y+1}\)

\(=1\)

\(\Rightarrow\frac{1}{x^2+2y^2+3}+\frac{1}{y^2+2z^2+3}+\frac{1}{z^2+2x^2+3}\le\frac{1}{2}\cdot1=\frac{1}{2}\left(đpcm\right)\)

Forever Miss You thiếu dấu "=" xảy ra khi nào:v

Dấu "=" xảy ra <=> x=y=z=1

Áp dụng bất đẳng thức Cauchy - Schwarz : \(\frac{a^2}{b}+\frac{c^2}{d}\ge\frac{\left(a+c\right)^2}{b+d}\) 

\(\frac{1}{x^4}+\frac{1}{y^4}=\frac{x^2}{x^6}+\frac{1^2}{y^4}\ge\frac{\left(x+1\right)^2}{x^6+y^4}\ge\frac{4x}{x^6+y^4}\)(\(\left(a+b\right)^2\ge4a\))

Tương tự: \(\frac{1}{y^4}+\frac{1}{z^4}\ge\frac{4y}{y^6+z^4};\frac{1}{z^4}+\frac{1}{x^4}\ge\frac{4z}{z^6+x^4}\)

\(\Rightarrow2.\left(\frac{1}{x^4}+\frac{1}{y^4}+\frac{1}{z^4}\right)\ge4\left(\frac{x}{x^6+y^4}+\frac{y}{y^6+z^4}+\frac{z}{z^6+x^4}\right)\)

\(\Rightarrow\frac{1}{x^4}+\frac{1}{y^4}+\frac{1}{z^4}\ge\frac{2x}{x^6+y^4}+\frac{2y}{y^6+z^4}+\frac{2z}{z^6+x^4}\)

Dấu "=" xảy ra khi và chỉ khi \(x=y=z=1\)

với x,y,z >0 áp dụng bđt cosi ta có:

\(x^6+y^4>=2\sqrt{x^6y^4}=2x^3y^2\Rightarrow\frac{2x}{x^6+y^4}< =\frac{2x}{2x^3y^2}=\frac{1}{x^2y^2}\)

\(y^6+z^4>=2\sqrt{y^6z^4}=2y^3z^2\Rightarrow\frac{2y}{y^6+z^4}< =\frac{2y}{2y^3z^2}=\frac{1}{y^2z^2}\)

\(z^6+x^4>=2\sqrt{z^6x^4}=2z^3x^2\Rightarrow\frac{2z}{z^6+x^4}< =\frac{2z}{2z^3x^2}=\frac{1}{z^2x^2}\)

\(\Rightarrow\frac{2x}{x^6+y^4}+\frac{2y}{y^6+z^4}+\frac{2z}{z^6+x^4}< =\frac{1}{x^2y^2}+\frac{1}{y^2z^2}+\frac{1}{z^2x^2}\left(1\right)\)

với x,y,z>0 áp dụng bđt cosi ta có:

\(\frac{1}{x^4}+\frac{1}{y^4}>=2\sqrt{\frac{1}{x^4}\cdot\frac{1}{y^4}}=\frac{2}{x^2y^2}\)

\(\frac{1}{y^4}+\frac{1}{z^4}>=2\sqrt{\frac{1}{y^4}\cdot\frac{1}{z^4}}=\frac{2}{y^2z^2}\)

\(\frac{1}{x^4}+\frac{1}{z^4}>=2\sqrt{\frac{1}{x^4}\cdot\frac{1}{z^4}}=\frac{2}{x^2z^2}\)

\(\Rightarrow\frac{2}{x^4}+\frac{2}{y^4}+\frac{2}{z^4}>=\frac{2}{x^2y^2}+\frac{2}{y^2z^2}+\frac{2}{x^2z^2}\Rightarrow\frac{1}{x^4}+\frac{1}{y^4}+\frac{1}{z^4}>=\frac{1}{x^2y^2}+\frac{1}{y^2z^2}+\frac{1}{x^2z^2}\)

\(\Rightarrow\frac{1}{x^2y^2}+\frac{1}{y^2z^2}+\frac{1}{x^2z^2}< =\frac{1}{x^4}+\frac{1}{y^4}+\frac{1}{z^4}\left(2\right)\)

từ \(\left(1\right)\left(2\right)\Rightarrow\frac{2x}{x^6+y^4}+\frac{2x}{y^6+z^4}+\frac{2x}{z^6+x^4}< =\frac{1}{x^4}+\frac{1}{y^4}+\frac{1}{z^4}\)(đpcm)

dấu = xảy ra khi x=y=z=1

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