bạn có thể tham khảo bài giải của c trong câu hỏi tương tự 

đặt a = 2x + y + z; b = 2y + z + x; c = 2z + x + y (a; b ; c > 0)

=> a + b + c = 4.(x+ y + z) => x + y + z = (a+ b+ c) / 4

=> x = a - (x+ y + z) = a - (a+ b + c) / 4 

y = b - (x + y + z) = b - (a+b+c) / 4

z = c - (x+y + z) = c - (a+b+c)/ 4 

Khi đó :  \(VT=1-\frac{a+b+c}{4a}+1-\frac{a+b+c}{4b}+1-\frac{a+b+c}{4c}\)

\(VT=3-\left(\frac{a+b+c}{4a}+\frac{a+b+c}{4b}+\frac{a+b+c}{4c}\right)=3-\frac{1}{4}.\left(a+b+c\right).\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(VT=3-\frac{1}{4}.\left(1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\right)=3-\frac{1}{4}.\left(3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right)\)

Với a, b > 0 ta có: a/b + b/ a > = 2

=> \(\frac{1}{4}.\left(3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right)\ge\frac{1}{4}.\left(3+2+2+2\right)=\frac{9}{4}\)

=> \(VT\le3-\frac{9}{4}=\frac{3}{4}\)

Dấu = xảy ra khi a= b = c => x = y = z 

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

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

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

Cộng theo vế:

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

Đặt \(\hept{\begin{cases}2x+y+z=a\\2y+z+x=b\\2z+x+y=c\end{cases}}\Rightarrow a+b+c=4\left(x+y+z\right)=\)

\(4\left(a-x\right)=4\left(b-y\right)=4\left(c-z\right)\Rightarrow\hept{\begin{cases}4x=3a-b-c\\4y=3b-c-a\\4z=3c-a-b\end{cases}}\)

Lúc đó thì \(4VT=\frac{3a-b-c}{a}+\frac{3b-c-a}{b}+\frac{3c-a-b}{c}\)

\(=3-\frac{b}{a}-\frac{c}{a}+3-\frac{c}{b}-\frac{a}{b}+3-\frac{a}{c}-\frac{b}{c}\)

\(=9-\left(\frac{a}{b}+\frac{b}{a}\right)-\left(\frac{b}{c}+\frac{c}{b}\right)-\left(\frac{c}{a}+\frac{a}{c}\right)\le3\)

\(\Rightarrow VT\le\frac{3}{4}\)

Đẳng thức xảy ra khi a = b = c hay x = y = z

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

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

Áp dụng tính chất : 1/a+b < = 1/4.(1/a+1/b) thì :

x/2x+y+z = x.(1/2x+y+z) = x.[1/(x+y)+(x+z)] < = x/4.(1/x+y + 1/x+z)

Tương tự : ..........

=> x/2x+y+z + y/x+2y+z + z/x+y+2z < = 1/4.(x/x+y + x/x+z + y/y+x + y/y+z + z/z+x + z/x+y )

                                                         = 1/4. [ ( x/x+y + y/x+y ) + ( y/y+z + z/z+y ) + ( z/z+x + x/x+z )

                                                         = 1/4.(1+1+1) = 3/4

Dấu "=" xảy ra <=> x=y=z

Vậy ..........

Tk mk nha

Đặt BT là P:

\(\text{P}=\frac{x}{\left(2x+y+z\right)}-1+\frac{y}{2y+z+x}-1+\frac{z}{\left(2z+x+y\right)}-1+3\)

\(\text{P}=-\frac{\left(x+y+z\right)}{\left(2x+y-z\right)}-\frac{\left(x+y+z\right)}{\left(2y+z+x\right)}-\frac{\left(x+y+z\right)}{\left(2z+x+y\right)}+3\)

\(\text{P}=-\left(x+y+z\right).\left[\frac{1}{\left(2x+y+z\right)}+\frac{1}{\left(2y+z+x\right)}+\frac{1}{\left(2z+x+y\right)}\right]+3\)

Co-si 3 số, ta có:

\(2x+y+z+2y+z+x+2z+x+y\ge3.\sqrt[3]{\left(2x+y+z\right)\left(2y+z+x\right)\left(2z+x+y\right)}\)

\(\Rightarrow4\left(x+y+z\right)\ge3.\sqrt[3]{\left(2x+y+z\right)\left(2y+z+x\right)\left(2z+x+y\right)}\)(1)

Co-si tiếp cho 3 số, ta có:

\(\frac{1}{\left(2x+y+z\right)}+\frac{1}{\left(2y+z+x\right)}+\frac{1}{\left(2z+x+y\right)}\ge3.\sqrt[3]{\frac{1}{\left(2x+y+z\right)}+\frac{1}{\left(2y+z+x\right)}+\frac{1}{\left(2z+x+y\right)}}\)(2)

Lấy (1) và (2) ta có: \(4\left(x+y+z\right)\left[\frac{1}{\left(2x+y+z\right)}+\frac{1}{\left(2y+z+x\right)}+\frac{1}{\left(2z+x+y\right)}\right]\ge9\)

\(\Rightarrow-\left(x+y+z\right).\left[\frac{1}{\left(2x+y+z\right)}+\frac{1}{\left(2y+z+x\right)}+\frac{1}{\left(2z+x+y\right)}\right]\le-\frac{9}{4}\)

Thay P, ta có:

\(\text{P}\le-\frac{9}{3}+3=\frac{3}{4}\left(ĐPCM\right)\)

Dấu "=" xảy ra khi x = y = z.

trả lời

x=y=z

hokt tốt

Áp dụng công thức \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\left(x,y>0\right)\)

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

\(\frac{1}{y+z}\le\frac{1}{4y}+\frac{1}{4z}\)

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

Tương tự \(\hept{\begin{cases}\frac{1}{x+2y+z}\le\frac{1}{4}\left(\frac{1}{4x}+\frac{1}{2y}+\frac{1}{4z}\right)\left(2\right)\\\frac{1}{x+y+2z}\le\frac{1}{4}\left(\frac{1}{4x}+\frac{1}{4y}+\frac{1}{2z}\right)\left(3\right)\end{cases}}\)

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

=> \(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le1\)

Dấu "=" xảy ra <=> \(x=y=z=\frac{3}{4}\)