\(\frac{a^2}{b^2}+\frac{b^2}{c^2}\ge2\sqrt{\frac{a^2b^2}{b^2c^2}}\ge\frac{2a}{c}\) ; \(\frac{a^2}{b^2}+\frac{c^2}{a^2}\ge\frac{2c}{b}\) ; \(\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{2b}{a}\)

Cộng vế với vế ta có đpcm

Dấu "=" xảy ra khi \(a=b=c\)

2. \(\frac{bc}{a}+\frac{ac}{b}\ge2\sqrt{\frac{bc.ac}{ab}}=2c\) ; \(\frac{ac}{b}+\frac{ab}{c}\ge2a\) ; \(\frac{bc}{a}+\frac{ab}{c}\ge2b\)

Cộng vế với vế ta có đpcm

Dấu "=" xảy ra khi \(a=b=c\)

Bài gắt quá, em cày mãi không ra:( nào là phân tích vế phải,sos từm lưm... Cuối cùng chuyển vế cho gọn:v Nhưng mà em ko chắc :((

BĐT \(\Leftrightarrow\Sigma_{cyc}\left(\frac{a^2}{b^2+c^2}-\frac{a}{b+c}\right)\ge0\)\(\Leftrightarrow\Sigma_{cyc}\frac{a^2b+a^2c-ab^2+ac^2}{\left(b^2+c^2\right)\left(b+c\right)}\ge0\)

\(\Leftrightarrow\Sigma_{cyc}\frac{ab\left(a-b\right)-ac\left(c-a\right)}{\left(b^2+c^2\right)\left(b+c\right)}\ge0\)\(\Leftrightarrow\Sigma_{cyc}\left[\frac{ab\left(a-b\right)}{\left(b^2+c^2\right)\left(b+c\right)}-\frac{ab\left(a-b\right)}{\left(c^2+a^2\right)\left(c+a\right)}\right]\ge0\)

\(\Leftrightarrow\Sigma_{cyc}ab\left(a-b\right)\left[\frac{\left(c^2+a^2\right)\left(c+a\right)-\left(b^2+c^2\right)\left(b+c\right)}{\left(b^2+c^2\right)\left(c^2+a^2\right)\left(b+c\right)\left(c+a\right)}\right]\ge0\)

\(\Leftrightarrow\Sigma_{cyc}ab\left(a-b\right)\left[\frac{\left(a-b\right)\left(a^2+b^2+c^2+ab+bc+ca\right)}{\left(b^2+c^2\right)\left(c^2+a^2\right)\left(b+c\right)\left(c+a\right)}\right]\ge0\)

\(\Leftrightarrow\left(a^2+b^2+c^2+ab+bc+ca\right).\Sigma_{cyc}\frac{ab\left(a-b\right)^2}{\left(b^2+c^2\right)\left(c^2+a^2\right)\left(b+c\right)\left(c+a\right)}\ge0\) (đúng)

Đẳng thức xảy ra khi a = b = c

áp dụng bất đẳng thức bu nhi a 

ta có \(3\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\ge\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\)

lại có  a/b+b/c+c/a \(\ge\)3 (bđt cauchy)

nhân từng vế ta có \(3\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\left(\frac{a}{b}+\frac{b}{a}+\frac{a}{c}\right)\ge3\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\)

suy ra đpcm

P/s: Không biết cách này có đúng không?

Chuyển vế qua và đặt thừa số chung,ta cần chứng minh:

\(a^2\left(\frac{1}{b+c}-\frac{1}{c+a}\right)+b^2\left(\frac{1}{a+c}-\frac{1}{a+b}\right)+c^2\left(\frac{1}{a+b}-\frac{1}{b+c}\right)\ge0\)

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

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

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

\(\Leftrightarrow a^2\left(a^2-b^2\right)+b^2\left(b^2-c^2\right)+c^2\left(c^2-a^2\right)\ge0\)

\(\Leftrightarrow a^4+b^4+c^4\ge a^2b^2+b^2c^2+c^2a^2\)

Đặt \(\left(a^2;b^2;c^2\right)\rightarrow\left(x;y;z\right)\).Ta cần chứng minh:

\(x^2+y^2+z^2\ge xy+yz+zx\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\) (đúng)

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

Ta có: x² – x – 12 = x² – x – 16 + 4

= (x² – 16) – (x – 4)

= (x – 4).(x + 4) – (x – 4)

= (x – 4).(x + 4 – 1)

= (x – 4).(x + 3)

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

\(\Rightarrow xyz=\frac{a}{b}.\frac{b}{c}.\frac{c}{a}=1\)

Bất đẳng thức đã cho tương đương với: \(\Leftrightarrow x^2+y^2+z^2\ge\frac{z}{x}+\frac{1}{y}+\frac{1}{z}\)

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

\(\Leftrightarrow2.\left(x^2+y^2+z^2\right)-2.\left(xy+yz+zx\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\left(\forall x;y;z\right)\)

Dấu "=" xảy ra khi \(\Leftrightarrow x=y=z\Rightarrow a=b=c\left(đpcm\right)\) 

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

\(\Leftrightarrow\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)\ge\frac{9}{2}\)

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge\frac{9}{2}\)

\(VT=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge\frac{\left(a+b+c\right).9}{2\left(a+b+c\right)}=\frac{9}{2}\) (đpcm)

Dấu "\("="\) xảy ra khi \(a=b=c\)

Câu b : \(VT=\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\left(đpcm\right)\)

Dấu = xảy ra khi a=b=c