where is "c" ?

a: Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

nên \(\dfrac{a}{c}=\dfrac{b}{d}\)

d: Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

nên \(\dfrac{a}{c}=\dfrac{b}{d}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)

hay \(\dfrac{a}{b}=\dfrac{a+c}{b+d}\)

Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

nên \(\dfrac{a}{c}=\dfrac{b}{d}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\)

hay \(\dfrac{a}{c}=\dfrac{a-b}{c-d}\) 

\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+c}{a+b+c}+\dfrac{a+b}{a+b+c}+\dfrac{b+c}{a+b+c}=2\) (1)

\(VP=\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}=\dfrac{a}{\sqrt{a\left(b+c\right)}}+\dfrac{b}{\sqrt{b\left(c+a\right)}}+\dfrac{c}{\sqrt{c\left(a+b\right)}}\)

\(VP\ge\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=2\) (2)

(1);(2) \(\Rightarrow VT< VP\)

\(a,\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\\ b,\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\\ \Leftrightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)

a) \(\dfrac{a}{c}=\dfrac{a+b}{c+d}\)

=> a(c + d) = c(a + b)

=> ac + ad = ac + bc

=> ad = bc \(\Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\)

b) \(\dfrac{b}{d}=\dfrac{a-b}{c-d}\)

=> b(c - d) = d(a - b)

=> bc - bd = ad - bd

=> bc = ad \(\Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\)

Ta có \(\dfrac{a^2}{b^2}+1\ge2.\dfrac{a}{b}\)

Lập 2 BĐT tương tự rồi cộng theo vế, ta được:

\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}+3\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\) (*)

Mà ta lại có \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge3\sqrt[3]{\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{a}}=3\)

\(\Leftrightarrow-3\ge-\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\) (**)

Cộng theo vế (*) và (**), ta được đpcm. 

Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

\(A=\dfrac{a}{b+c}+\dfrac{b+c}{a}+\dfrac{b}{c+a}+\dfrac{c+a}{b}+\dfrac{c}{a+b}+\dfrac{a+b}{c}\)

\(A=\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)+\left(\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\right)\)

\(A\ge\dfrac{3}{2}+\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\) (bất đẳng thức Nesbit)

\(A\ge\dfrac{3}{2}+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{c}\)

\(A\ge\dfrac{3}{2}+\left(\dfrac{b}{a}+\dfrac{a}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)

Áp dụng bất đẳng thức AM-GM cho 2 số dương ta có:

\(A\ge\dfrac{3}{2}+2\sqrt{\dfrac{ab}{ab}}+2\sqrt{\dfrac{ac}{ac}}+2\sqrt{\dfrac{bc}{bc}}\)

\(A\ge\dfrac{3}{2}+2+2+2=\dfrac{15}{2}\left(đpcm\right)\)

Dấu"=" xảy ra khi: \(a=b=c\)

Hình như thế này mới đúng chứ \(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\)

Áp dụng BĐT Cosi:

\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2.\dfrac{a}{c};\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2.\dfrac{b}{a};\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2.\dfrac{c}{b}\)

\(\Rightarrow2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\right)\)

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

Đẳng thức xảy ra khi \(a=b=c>0\)