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

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

Có : (a-b)^2>=0

<=> a^2+b^2-2ab >=0

<=>a^2+b^2 >= 2ab

<=>a^2+b^2+2ab >= 4ab

<=> (a+b)^2 >= 4ab

Với a,b >0 thì chia cả 2 vế cho (a+b).ab thì :

a+b/ab >= 4/a+b

<=>4/a+b <= 1/a+1/b

<=> 1/a+b <= 1/4.(1/a+1/b)         ( với mọi a,b > 0 )

Áp dụng bđt trên cho x;y;z > 0 thì : x/2x+y+z = x. 1/(x+y)+(z+x) <= x/4 .( 1/x+y+1/x+z) = x/4.(x+y) + x/4.(x+z)

Tương tự : y/x+2y+z <= y/4.(y+x) + y/4.(y+z)

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

=> VT <= [ x/4.(x+y) + y/4.(y+x) ] + [ y/4.(y+z) + z/4.(z+y) ] + [ z/4.(z+x) + x/4.(x+z) ] = 1/4 + 1/4 + 1/4 = 3/4

=> ĐPCM

Dấu "=" xảy ra <=> x=y=z > 0 

k mk nha

áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) với mọi a,b >0 

Thì \(\frac{x}{x+y}+\frac{x}{x+z}\ge\frac{4x}{2x+y+z}\) 

Tương tự thì đpcm 

Cách này nhanh này thành đơ

hỏi chi hỏi lắm rứa

Bài của lớp 7 ghê vậy!!

Áp dụng bất đẳng thức Cauchy cho 3 số dương x,y,z

ta có bổ đề \((a+b+c)({1\over a}+{1\over b}+{1\over c})\)  >  9

Áp dụng vào ta có

\(D*({2x+y+z\over x}+{2y+x+z\over y}+{2z+y+x\over z})\)  >9(1)

Ta có \({2x+y+z\over x}+{2y+x+z\over y}+{2z+y+x\over z}\) =\(2+{y+z\over x}+2+{z+x\over y}+2+{y+x\over z}\)=\(6-3+{y+z\over x}+1+{z+x\over y}+1+{y+x\over z}+1\)=\(3+{x+y+z\over x}+{y+x+z\over y}+{z+y+x\over z}\)=\(3+(x+y+z)({1\over x}+{1\over y}+{1\over z})\)  >  3+9=12

thay vào(1)

Ta có \(D \) <  \({9\over 12}\)=\({3\over 4}\) 

Dấu "=" xảy ra khi x=y=z 

=> ĐPCM

áp dụng bất đẳng thức phụ : \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

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

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

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

cộng vế theo vế

\(\frac{x}{2x+y+z}+\frac{y}{2y+z+x}+\frac{z}{2z+x+y}\le\frac{1}{4}\left(\frac{x+y}{x+y}+\frac{y+z}{y+z}+\frac{z+x}{z+x}\right)=\frac{1}{4}\cdot3=\frac{3}{4}\)(đpcm)

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

\(tươngtự:\frac{y}{2y+z+x}\le\frac{1}{16}\left(2+\frac{y}{z}+\frac{y}{x}\right);\frac{z}{2z+x+y}\le\frac{1}{16}\left(2+\frac{z}{x}+\frac{z}{y}\right).\text{Cộng vế theo vế ta được:}\frac{x}{2x+y+z}+\frac{y}{2y+z+x}+\frac{z}{2z+y+x}\le\frac{1}{16}\left(2+2+2+\frac{x}{y}+\frac{y}{x}+\frac{z}{x}+\frac{x}{z}+\frac{y}{z}+\frac{z}{y}\right)=\frac{1}{16}\left[6+\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{z}{x}+\frac{x}{z}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)\right]\ge\frac{1}{16}\left(6+2\sqrt{\frac{xy}{xy}}+2\sqrt{\frac{xz}{xz}}+2\sqrt{\frac{yz}{yz}}\right)=\)

\(=\frac{12}{16}=\frac{3}{4}\Rightarrow\frac{x}{2x+y+z}+\frac{y}{2y+z+x}+\frac{z}{2z+x+y}\le\frac{3}{4}\left(\text{đpcm}\right)\)

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

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