\(\frac{a}{2b}+\frac{b}{2a}\ge1\)

\(\frac{2a^2}{4ba}+\frac{2b^2}{4ab}\ge1\)

\(2a^2+2b^2\ge1\)( do số bình phương luôn luôn lớn hơn 0)

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

Tương tự, ta có: \(\frac{\left(2-a\right)\left(c-a\right)}{2b+ca}=\frac{c}{a+b}-\frac{a}{a+b};\frac{\left(2-b\right)\left(a-b\right)}{2c+ab}=\frac{a}{b+c}-\frac{b}{b+c}\)

\(\Rightarrow\)\(VT=\left(\frac{a}{b+c}-\frac{a}{a+b}\right)+\left(\frac{b}{c+a}-\frac{b}{b+c}\right)+\left(\frac{c}{a+b}-\frac{c}{c+a}\right)\)

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

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

\(=\frac{\left(a^3+b^3+c^3\right)-\left(a^2b+b^2c+c^2a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge\frac{\left(a^3+b^3+c^3\right)-\left(a^3+b^3+c^3\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{2}{3}\)

cái bđt \(a^3+b^3+c^3\ge a^2b+b^2c+c^2a\) cô Chi có làm r ib mk gửi link 

Ta có : \(\frac{a}{b}+\frac{b}{a}-2\)

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

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

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

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

\(=\frac{a\left(a-b\right)-b\left(a-b\right)}{ab}\)

\(=\frac{\left(a-b\right)\left(a-b\right)}{ab}\)

\(=\frac{\left(a-b\right)^2}{ab}\ge0\) với mọi \(a;b\inℕ^∗\)

\(\Rightarrow\frac{a}{b}+\frac{b}{a}-2\ge0\) với mọi \(a;b\inℕ^∗\)

\(\Rightarrow\frac{a}{b}+\frac{b}{a}\ge2\) với mọi \(a;b\inℕ^∗\) 

Ta có\(\frac{a}{b}+\frac{b}{a}-2\)

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

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

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

\(=\frac{a\left(a-b\right)-b\left(a-b\right)}{ab}\)

\(=\frac{\left(a-b\right)\left(a-b\right)}{ab}\)

\(=\frac{\left(a-b\right)^2}{ab}\ge0\text{ với mọi a;b \inℕ^∗}\)

\(\Rightarrow\frac{a}{b}+\frac{b}{a}-2\ge0\text{ với mọi a;b\inℕ^∗}\)

\(\Rightarrow\frac{a}{b}+\frac{b}{a}\ge2\text{ với mọi a;b \inℕ^∗}\)

Học tốt

Ta có:Xét hiệu \(\frac{a}{b}+\frac{b}{a}-2\)

\(\Rightarrow\frac{a}{b}+\frac{b}{a}-2=\frac{a^2-2ab+b^2}{ab}=\frac{\left(a-b\right)^2}{ab}\ge0\)(Vì\(a,b\inℕ^∗\))

\(\Rightarrow\frac{a}{b}+\frac{b}{a}\ge2\)(Đấu "=" xảy ra khi và chỉ khi a=b)(đpcm)

giả sử a\(\ge\)b không làm mất đi tính chất tổng quát của bài.

\(\Rightarrow\)a = m  + b [ m \(\ge\)0]

ta có :

\(\frac{a}{b}+\frac{b}{a}=\frac{b+m}{b}\)\(\frac{b}{b+m}=1+\frac{m+b}{b+m}\)\(=1+1=2\)

\(vậy\)\(\frac{a}{b}+\frac{b}{a}\ge2(ĐPCM)\)

giả sử a \(\ge\)b \(\Rightarrow\)a = b + m ( m \(\ge\)0 )

do đó : \(\frac{a}{b}+\frac{b}{a}=\frac{b+m}{b}+\frac{b}{b+m}\)

\(=1+\frac{m}{b}+\frac{b}{b+m}\ge1+\frac{m}{b+m}+\frac{b}{b+m}=1+\frac{m+b}{b+m}=2\)

Vậy \(\frac{a}{b}+\frac{b}{a}\ge2\)( a,b thuộc N* )

Dấu " = " xảy ra khi a = b