a) Đặt: \(b+c=x;c+a=y;a+b=z\)

Có: \(x+y-z=b+c+c+a-a-b=2c\)

=> \(c=\frac{x+y-z}{2}\)

Tương tự ta cũng có:

\(a=\frac{y+z-x}{2};b=\frac{x+z-y}{2}\)

Có: \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

=\(\frac{y+z-x}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}\)

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

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

Áp dụng bđt cô si ta có:

\(\frac{y}{x}+\frac{x}{y}\ge2;\frac{z}{x}+\frac{x}{z}\ge2;\frac{z}{y}+\frac{y}{z}\ge2\)

=> \(\left(1\right)\ge\frac{1}{2}\left(2+2+2-3\right)=\frac{3}{2}\)

Vậy \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)

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

VÌ: \(\left[2a-\left(b+c\right)\right]^2\ge0\)

=> \(\left(2a\right)^2+\left(b+c\right)^2\ge4a\left(b+c\right)\)

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

Hay: \(\frac{a^2}{b+c}+\frac{b+c}{4}\ge a\Rightarrow\frac{a^2}{b+c}\ge a-\frac{b+c}{4}\) (2)

Tương tự ta cũng có: \(\frac{b^2}{c+a}\ge b-\frac{c+a}{4}\) (3)

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

Cộng vế với vế (2);(3);(4) ta có:

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

xin phép làm lại :3

a) \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

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

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

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

\(=\frac{1}{2}\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)

\(\ge\frac{1}{2}\cdot3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\cdot\frac{3}{\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}-3=\frac{3}{2}\)( đpcm )

Dấu "=" xảy ra <=> a=b=c

Ta có a/b =b/c 

=> a^2/b^2=a/b.a/b= a/b.b/c=a/c(1)

Lại có a/b=b/c

=> a^2/b^2=b^2/c^2=a^2+b^2  /  b^2+c^2 (t/c dãy tỉ số = nhau) (2)

Từ (1),(2) => a/c=a^2+b^2  /  b^2+c^2

Ta có \(\frac{a}{b}=\frac{b}{c}\)=> \(\left(\frac{a}{b}\right)^2=\left(\frac{b}{c}\right)^2\)

                             => \(\frac{a^2}{b^2}=\frac{b^2}{c^2}=\frac{a^2+b^2}{b^2+c^2}\)mà \(\frac{a}{b}=\frac{b}{c}\)
                            => \(\frac{a^2+b^2}{b^2+c^2}=\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}=\frac{a}{c}\)

Ta có : \(\frac{a}{b}=\frac{b}{c}\)

\(\Rightarrow\frac{a^2}{b^2}=\frac{b^2}{c^2}\)

Áp dung tính chất của dãy tỉ bằng nhau , ta có :

\(\frac{a^2}{b^2}=\frac{b^2}{c^2}=\frac{a^2+b^2}{b^2+c^2}=\frac{a}{c}.\frac{a}{c}=\frac{a}{c}.\frac{b}{c}=\frac{a}{c}\)( điều phải chứng minh )

Vậy ...............

bài này chắc có câu a đúng ko

ta có \(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}=\frac{a}{c}=\frac{c}{b}=\frac{b}{a}\)

\(\Leftrightarrow a^4c^2+b^4a^2+c^4b^2=abc\left(a^2c+c^2a+b^2c\right)\)

đặt \(x=a^2c;y=b^2a;z=c^2b\)ta được

\(x^2+y^2+z^2=xy+yz+zx\)

áp dụng kết quả của câu a ta đc

\(\left(x-y\right)^2+\left(y-2\right)^2+\left(z-x\right)^2=0=>x=y=z\)

\(=>a^2c=b^2a=c^2b=>ac=b^2;bc=a^2;ab=c^2\)

=>a=b=c(dpcm)

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

Đặt \(\frac{a}{b}=x;\frac{b}{c}=y;\frac{c}{a}=z\)

Khi đó:\(x^2+y^2+z^2=xy+yz+zx\)

\(\Leftrightarrow2\left(x^2+y^2+z^2\right)=2\left(xy+yz+zx\right)\)

\(\Leftrightarrow2\left(x^2+y^2+z^2\right)-2\left(xy+yz+zx\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

Mà \(\left(x-y\right)^2\ge0;\left(y-z\right)^2\ge0;\left(z-x\right)^2\ge0\)

\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

Dấu "=" xảy ra tại x=y=z hay a=b=c

Suy ra điều fải chứng minh

Ta có: 

\(\frac{\overline{ab}}{\overline{bc}}=\frac{b}{c}\)

<=> \(\frac{a.10+b}{b.10+c}=\frac{b}{c}\)

Áp dụng dãy tỉ số bằng nhau ta có: 

\(\frac{a.10+b}{b.10+c}=\frac{b}{c}=\frac{10a+b-b}{10b+c-c}=\frac{10a}{10b}=\frac{a}{b}\)

=> \(\frac{b}{c}=\frac{a}{b}\Rightarrow b^2=ac\)

khi đó: \(\frac{a^2+b^2}{b^2+c^2}=\frac{a^2+ac}{ac+c^2}=\frac{a\left(a+c\right)}{c\left(a+c\right)}=\frac{a}{c}\)

Vậy:...