Ta có: S = (a/a+b+c +b/a+b+c +c/a+b+c) + (a+b+c /a + a+b+c /b + a+b+c /c) -a/b -a/c -b/a -b/c -c/a -c/b

              = a+b+c/ a+b+c + 1+ b/a +c/a +a/b +1 +c/b +a/c +b/c +1 -a/b -a/c -b/a -b/c -c/a- c/b

              = 1+1+1+1 (vì a+b+c khác 0)

              = 4

Vậy S = 4

Bài này bạn chỉ cần tách đơn giản vậy thôi.Chúc bạn học tốt.

m nhanh đó Phạm Trọng An Nam à

a) \(A=\left(1+\frac{b^2+c^2-a^2}{2bc}\right).\frac{1+\frac{a}{b+c}}{1-\frac{a}{b+c}}.\frac{b^2+c^2-\left(b-c\right)^2}{a+b+c}\)

\(=\frac{2bc+b^2+c^2-a^2}{2bc}.\frac{\frac{a+b+c}{b+c}}{\frac{b+c-a}{b+c}}.\frac{b^2+c^2-b^2+2bc-c^2}{a+b+c}\)

\(=\frac{\left(b+c+a\right)\left(b+c-a\right)}{2bc}.\frac{a+b+c}{b+c-a}.\frac{2bc}{a+b+c}\)

\(=a+b+c\)

b) \(B=\frac{\frac{3a}{a+b}}{\frac{2a}{a^2-2ab+b^2}}\)\(=\frac{3a}{a+b}.\frac{\left(a-b\right)^2}{2a}=\frac{3\left(a-b\right)^2}{2\left(a+b\right)}\)

c) \(C=\frac{\frac{a}{b}+\frac{b}{a}}{\frac{a}{b}-\frac{b}{a}}:\frac{\frac{a^2}{b^2}-\frac{b^2}{a^2}}{\left(\frac{1}{a}+\frac{1}{b}\right)^2}\)

\(=\frac{\frac{a^2+b^2}{ab}}{\frac{a^2-b^2}{ab}}:\frac{\frac{a^4-b^4}{a^2b^2}}{\frac{\left(a+b\right)^2}{a^2b^2}}\)

\(=\frac{a^2+b^2}{a^2-b^2}.\frac{\left(a+b\right)^2}{a^4-b^4}\)

\(=\frac{\left(a^2+b^2\right)\left(a+b\right)^2}{\left(a+b\right)\left(a-b\right)\left(a^2+b^2\right)\left(a+b\right)\left(a-b\right)}\)

\(=\frac{1}{\left(a-b\right)^2}\)

Dảnh àk =))

Cứ đăng đi - úng hộ ^^

Ta chứng minh BĐT sau với các số dương:

\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)

Thật vậy, BĐT tương đương: \(\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)

Áp dụng:

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) ; \(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\) ; \(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\)

Cộng vế với vế:

\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\)

\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\)

b.

Ta có:

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Rightarrow\dfrac{3}{a}+\dfrac{3}{b}\ge\dfrac{12}{a+b}\) (1)

\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\Rightarrow\dfrac{2}{b}+\dfrac{2}{c}\ge\dfrac{8}{b+c}\) (2)

\(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\) (3)

Cộng vế với vế (1); (2) và (3):

\(\dfrac{4}{a}+\dfrac{5}{b}+\dfrac{3}{c}\ge4\left(\dfrac{3}{a+b}+\dfrac{2}{b+c}+\dfrac{1}{c+a}\right)\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

Lần sau đăng ít một thôi toàn bài dài :v, ko phải ko làm mà là ngại làm

a)Áp dụng BĐT Cauchy-Schwarz ta có:

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

Tương tự cho 2 BĐT còn lại ta cũng có:

\(\frac{b}{a+2b+c}\le\frac{1}{4}\left(\frac{b}{a+b}+\frac{b}{b+c}\right);\frac{c}{a+b+2c}\le\frac{1}{4}\left(\frac{c}{a+c}+\frac{c}{b+c}\right)\)

Cộng theo vế 3 BĐT trên ta có:

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

Xảy ra khi \(a=b=c\)

b)Đặt \(THANG=abc\left(a^2+bc\right)\left(b^2+ac\right)\left(c^2+ab\right)>0\)

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{b+c}{a^2+bc}-\frac{c+a}{b^2+ac}-\frac{a+b}{a^2+ab}\)

\(=\frac{a^4b^4+b^4c^4+c^4a^4-a^4b^2c^2-b^4c^2a^2-c^4a^2b^2}{THANG}\)

\(=\frac{\left(a^2b^2-b^2c^2\right)^2+\left(b^2c^2-c^2a^2\right)+\left(c^2a^2-a^2b^2\right)^2}{2THANG}\ge0\) (Đúng)

Xảy ra khi \(a=b=c\)

c)Ta có:\(\frac{a^2}{b^2+c^2}-\frac{a}{b+c}=\frac{ab\left(a-b\right)+ac\left(a-c\right)}{\left(b+c\right)\left(b^2+c^2\right)}\)

Và \(\frac{b^2}{c^2+a^2}-\frac{b}{c+a}=\frac{bc\left(b-c\right)+ab\left(b-a\right)}{\left(c+a\right)\left(c^2+a^2\right)}\)

\(\frac{c^2}{a^2+b^2}-\frac{c}{a+b}=\frac{ac\left(c-a\right)+bc\left(c-b\right)}{\left(b+a\right)\left(b^2+a^2\right)}\)

Cộng theo vế 3 đăng thức trên ta có:

\(VT-VP=Σ\left[\frac{ab\left(a-b\right)}{\left(b+c\right)\left(b^2+c^2\right)}-\frac{ab\left(a-b\right)}{\left(a+c\right)\left(a^2+c^2\right)}\right]\)

\(=\left(a^2+b^2+c^2+ab+bc+ca\right)\cdotΣ\frac{ab\left(a-b\right)^2}{\left(b+c\right)\left(c+a\right)\left(b^2+c^2\right)\left(c^2+a^2\right)}\ge0\)

2 bài cuối full quy đồng mệt thật :v

trừ mỗi tỉ lệ cho 1 ta được:

\(\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)

\(\Rightarrow\frac{2a+b+c+d}{a}-\frac{a}{a}=\frac{a+2b+c+d}{b}-\frac{b}{b}=\frac{a+b+2c+d}{c}-\frac{c}{c}=\frac{a+b+c+2d}{d}-\frac{d}{d}\)

\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)

+Nếu a+b+c+d\(\ne\)0 thì a=b=c=d lúc đó 

M=1+1+1+1=4

+Nếu a+b+c+d=0 thì a+b=-(c+d);b+c=-(d+a);c+d=-(a+b);d+a=-(b+c) lúc đó:

M=(-1)+(-1)+(-1)+(-1)=-4

\(\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}=\frac{a+b+2c+d+a+b+c+2d}{c+d}=\frac{2a+2b+3c+3d}{c+d}\)

\(=\frac{2\left(a+b\right)}{c+d}+\frac{3\left(c+d\right)}{c+d}=2.\frac{a+b}{c+d}+3\)

\(\frac{2a+b+c+d}{a}=\frac{a+b+c+2d}{d}=\frac{2a+b+c+d+a+b+c+2d}{a+d}=\frac{3a+3d+2c+2b}{a+d}\)

\(=\frac{3\left(a+d\right)}{a+d}+\frac{2\left(b+c\right)}{a+d}=3+2.\frac{b+c}{a+d}\)

\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{2a+b+c+d+a+2b+c+d}{a+b}=\frac{3a+3b+2c+2d}{a+b}\)

\(=\frac{3\left(a+b\right)}{a+b}+\frac{2\left(c+d\right)}{a+b}=3+\frac{c+d}{a+b}.2\)

\(\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+2b+c+d+a+b+2c+d}{b+c}=\frac{3b+3c+2a+2d}{b+c}\)

\(=\frac{3\left(b+c\right)}{b+c}+\frac{2\left(a+d\right)}{b+c}=3+\frac{a+d}{b+c}.2\)

\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}=\frac{5\left(a+b+c+d\right)}{a+b+c+d}=5\)

\(\Rightarrow\frac{2a+b+c+d}{a}+\frac{a+2b+c+d}{b}+\frac{a+b+2c+d}{c}+\frac{a+b+c+2d}{d}=5.4=20\)

\(\Rightarrow3+\frac{a+b}{c+d}.2+3+\frac{b+c}{a+d}.2+3+\frac{c+d}{a+b}.2+3+\frac{d+a}{b+c}.2=20\)

\(\Rightarrow2.\left(\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\right)=20-3-3-3-3\)

\(\Rightarrow\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{b+a}+\frac{d+a}{b+c}=8:2=4\)

vậy \(\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=4\)

1) Theo bđt AM-GM,ta có: \(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2\sqrt{\frac{a^2}{b+c}.\frac{b+c}{4}}=a\)

Suy ra \(\frac{a^2}{b+c}\ge a-\frac{b+c}{4}\)

Thiết lập hai BĐT còn lại tương tự và cộng theo vế ta có đpcm

4/\(\frac{a^2}{b}+b\ge2\sqrt{\frac{a^2}{b}.b}=2a\Rightarrow\frac{a^2}{b}\ge2a-b\)

Thiết lập 2 BĐT còn lai5n tương tự,cộng theo vế ta có đpcm.

3/Theo BĐT AM-GM,ta có: \(ab\le\frac{\left(a+b\right)^2}{4}\Rightarrow\frac{ab}{a+b}\le\frac{\left(a+b\right)^2}{4\left(a+b\right)}=\frac{a+b}{4}\)

Thiết lập hai BĐT còn lại tương tự và cộng theo vế ta có đpcm

\(D=\left(\frac{a-b}{a^{\frac{3}{4}}+a^{\frac{1}{2}}.b^{\frac{1}{4}}}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}\right):\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)^{-1}\sqrt{\frac{a}{b}}\)

   \(=\left[\frac{a-b}{a^{\frac{1}{2}}\left(a^{\frac{1}{4}}+b^{\frac{1}{4}}\right)}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}\right]:\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)^{-1}\sqrt{\frac{b}{a}}\)

    \(=\frac{a-b-a+a^{\frac{1}{2}}.b^{\frac{1}{2}}}{a^{\frac{1}{2}}\left(a^{\frac{1}{4}}+b^{\frac{1}{4}}\right)}.\frac{1}{\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)}=\frac{b^{\frac{1}{2}}}{a^{\frac{1}{2}}}\frac{\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)}{\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)}\sqrt{\frac{a}{b}}.\sqrt{\frac{a}{b}}=1\)