Ta có :  \(\left(x+y\right)^2\ge4xy\Rightarrow\frac{x+y}{xy}\ge\frac{4}{x+y}\)

\(\Rightarrow\frac{4}{x+y}\le\frac{1}{x}+\frac{1}{y}\)

\(\Rightarrow\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)

Áp dụng ta có :

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

Tương tự : 

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

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

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

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

\(\Rightarrow4\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\le\frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}\)

\(\Rightarrow4\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\le\frac{a+c}{b}+\frac{a+b}{c}+\frac{b+c}{a}\)

\(\Rightarrow4\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\le\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\)

Dấu = xảy ra khi a=b=c

Áp dụng BĐT cô si ta có : 

\(\frac{b+c}{a}\ge4.\frac{a}{b+c}\)

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

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

\(\Rightarrow\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\ge4.\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\)

Dấu " = " xảy ra khi a=  b = c

chung minh gi va

ghi ra

Từ bất đẳng thức Cô si ta có:

\(4\left(ab+bc+ca\right)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\le\left[\frac{ab+bc+ca}{ca}+ca\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\right]^2\)

\(\Rightarrow\)Ta cần chứng minh:

\(\frac{ab+bc+ca}{ca}+ca\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\le\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\)

Vì vai trò của a, b, c trong bất đẳng thức như nhau, nên không mất tính tổng quát ta giả sử \(a\ge b\ge c\)nên bất đẳng thức cuối cùng đùng. Vậy bất đẳng thức được chứng minh.

sai r bạn ơi ko biết còn đòi

a)

Đặt   \(A=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)

\(\Rightarrow A=\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{c^2}{ac+bc}\)

Áp dụng BĐT Schwarz , ta có :

\(A\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}\)  (1)

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

\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2\ge0\)

\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ac\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac\ge3\left(ab+bc+ac\right)\)

\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)

\(\Leftrightarrow\frac{\left(a+b+c\right)^2}{ab+bc+ac}\ge3\)     (2)

Từ (1) và (2) , suy ra :  \(A\ge\frac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c\)

b)

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

 tại sao lại dc cái này bạn

\(\frac{\left(a+b\right)^2}{c}+\frac{\left(b+c\right)^2}{a}+\frac{\left(c+a\right)^2}{b}\ge\frac{\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]^2}{a+b+c}\)

BDDT Schawars :

\(\frac{x^2}{a}+\frac{y^2}{b}\ge\frac{\left(x+y\right)^2}{a+b}\) ( vs a,b,x,y dương )

\(\Leftrightarrow x^2b\left(a+b\right)+y^2a\left(a+b\right)=ab\left(x+y\right)^2\)

\(\Leftrightarrow x^2ab+x^2b^2+y^2a^2+y^2ab\ge x^2ab+2abxy+y^2ab\)

\(\Leftrightarrow x^2b^2-2abxy+y^2a^2\ge0\)

\(\Leftrightarrow\left(xb-ya\right)^2\ge0\) ( Luôn đúng )

''='' khi \(xb=ya\Leftrightarrow\frac{x}{a}=\frac{y}{b}\)

Áp dụng , ta có :

\(B=\frac{\left(a+b\right)^2}{c}+\frac{\left(b+c\right)^2}{a}+\frac{\left(c+a\right)^2}{b}\ge\frac{\left[\left(a+b\right)+\left(b+c\right)\right]^2}{a+c}+\frac{\left(c+a\right)^2}{b}\)

\(\Rightarrow B\ge\frac{\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]^2}{a+b+c}=\frac{\left[2\left(a+b+c\right)\right]^2}{a+b+c}=\frac{4\left(a+b+c\right)^2}{a+b+c}=4\left(a+b+c\right)\)

 Dấu "=" xảy ra khi \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\Rightarrow\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\Rightarrow a=b=c\)

Ta có: \(\frac{1}{a}+\frac{1}{c}=\frac{2}{b}\Leftrightarrow\frac{a+c}{ac}=\frac{2}{b}\Rightarrow b=\frac{2ac}{a+c}\)

Khi đó:

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

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

\(=\frac{a^2+3ac}{2a^2}+\frac{c^2+3ac}{2c^2}=\frac{a^2}{2a^2}+\frac{3ac}{2a^2}+\frac{c^2}{2c^2}+\frac{3ac}{2c^2}\)

\(=\frac{1}{2}+\frac{3c}{2a}+\frac{1}{2}+\frac{3a}{2c}=1+\frac{3}{2}\left(\frac{a}{c}+\frac{c}{a}\right)\)

\(\ge1+\frac{3}{2}\cdot2\sqrt{\frac{a}{c}\cdot\frac{c}{a}}=1+3=4\) (Cauchy)

Dấu "=" xảy ra khi: \(a=b=c\)