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

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

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

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

\(\Rightarrow VP>VT\) (đpcm)

Ta có: Theo bất đẳng thức cauchy schwarz và bất đẳng thức cauchy với a;b;c>0 ta có:

\(\dfrac{1}{a^2}+\dfrac{1}{a^2}=\dfrac{\left(\sqrt{a}\right)^2}{a^3}+\dfrac{1}{a^2}\ge\dfrac{\left(\sqrt{a}+1\right)^2}{a^3+a^2}\ge\dfrac{4\sqrt{a}}{a^3+a^2}\)(1)

Tương tự \(\dfrac{1}{b^2}+\dfrac{1}{b^2}\ge\dfrac{4\sqrt{b}}{b^3+b^2}\left(2\right);\dfrac{1}{c^2}+\dfrac{1}{c^2}\ge\dfrac{4\sqrt{c}}{c^3+c^2}\left(3\right)\)

Cộng từng vế (1) ;(2);(3) vế theo vế rồi chia hai vế cho 2 ta có đpcm

Áp dụng BĐT Cauchy schwarz kết hợp với AM-GM cho các số dương ta có :

\(\dfrac{1}{a^2}+\dfrac{1}{b^2}=\dfrac{a}{a^3}+\dfrac{1}{b^2}\ge\dfrac{\left(\sqrt{a}+1\right)^2}{a^3+b^2}\ge\dfrac{4\sqrt{a}}{a^3+b^2}\)

\(\dfrac{1}{b^2}+\dfrac{1}{c^2}=\dfrac{b}{b^3}+\dfrac{1}{c^2}\ge\dfrac{\left(\sqrt{b}+1\right)^2}{b^3+c^2}\ge\dfrac{4\sqrt{b}}{b^3+c^2}\)

\(\dfrac{1}{c^2}+\dfrac{1}{a^2}=\dfrac{c}{c^3}+\dfrac{1}{a^2}\ge\dfrac{\left(\sqrt{c}+1\right)^2}{c^3+a^2}\ge\dfrac{4\sqrt{c}}{c^3+a^2}\)

Cộng từng vế của BĐT ta được :

\(2\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\ge\dfrac{4\sqrt{a}}{a^3+b^2}+\dfrac{4\sqrt{b}}{b^3+c^2}+\dfrac{4\sqrt{c}}{c^3+a^2}\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{2\sqrt{a}}{a^3+b^2}+\dfrac{2\sqrt{b}}{b^3+c^2}+\dfrac{2\sqrt{c}}{c^3+a^2}\) ( đpcm )

Dấu \("="\) xảy ra khi \(a=b=c\)

ta có : \(\dfrac{1}{a^2}+\dfrac{1}{b^2}=\dfrac{a}{a^3}+\dfrac{1}{b^2}\ge\dfrac{\left(\sqrt{a}+1\right)^2}{a^3+b^2}=\dfrac{a^2+2\sqrt{a}+1}{a^3+b^2}\ge\dfrac{4\sqrt{a}}{a^3+b^2}\)

làm tương tự ta có : \(\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{4\sqrt{b}}{b^3+c^2}\) và \(\dfrac{1}{c^2}+\dfrac{1}{a^2}\ge\dfrac{4\sqrt{c}}{c^3+a^2}\)

cộng quế theo quế \(\Rightarrow\) (đpcm)

\(\frac{1}{a^2}+\frac{1}{b^2}=\frac{a}{a^3}+\frac{1}{b^2}\ge\frac{\left(\sqrt{a}+1\right)^2}{a^3+b^2}\ge\frac{4\sqrt{a}}{a^3+b^2}\)

Cứ tiếp tục như vậy ta sẽ có đpcm. dấu = xảy ra khi a=b=c=1

đề bài sai rồi bạn nhé check lại đi 

Sửa đề: \(\frac{a}{b}+\frac{a}{c}+\frac{c}{b}+\frac{c}{a}+\frac{b}{c}+\frac{b}{a}\ge\sqrt{2}\left(\Sigma\sqrt{\frac{1-a}{a}}\right)\)

or \(\Sigma\frac{b+c}{a}\ge\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}\)

Theo AM-GM:\(\frac{b+c}{a}\ge2\sqrt{\frac{2\left(b+c\right)}{a}}-2\)

Tương tự và cộng lại: \(VT\ge2\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}-6\)

Mà: \(\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}\ge3\sqrt[6]{\frac{8\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}}\ge6\)

Từ đó: \(VT\ge2\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}-\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}=VP\)

Done!

Vào ghé thăm nhà mình nhé: See method from solution! Cảm ơn bạn.

1,

\(\frac{a}{1+\frac{b}{a}}+\frac{b}{1+\frac{c}{b}}+\frac{c}{1+\frac{a}{c}}=\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\ge\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}=\frac{2}{2}=1\left(Q.E.D\right)\)

https://goo.gl/BjYiDy

1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)

\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\)  (1) 

áp dụng (x² +y² +z²)(m²+n²+p²) \(\ge\left(xm+yn+zp\right)^2\)

(2a²bc +b²c² + 2b²ac+a²c² + 2c²ab+a²b²). VT\(\ge\left(bc+ca+ab\right)^2\)   <=> (ab+bc+ca)². VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\)  ( vậy (1) đúng)

dấu '=' khi a=b=c

4b, \(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}=1-\frac{ab^2}{a^2+b^2}+1-\frac{bc^2}{b^2+c^2}+1-\frac{ca^2}{a^2+c^2}\)

\(\ge3-\frac{ab^2}{2ab}-\frac{bc^2}{2bc}-\frac{ca^2}{2ac}=3-\frac{\left(a+b+c\right)}{2}=\frac{3}{2}\)

4c, 

\(\frac{a+1}{b^2+1}+\frac{b+1}{c^2+1}+\frac{c+1}{a^2+1}=a+b+c-\frac{b^2}{b^2+1}-\frac{c^2}{c^2+1}-\frac{a^2}{a^2+1}+3--\frac{b^2}{b^2+1}-\frac{c^2}{c^2+1}-\frac{a^2}{a^2+1}\)\(\ge6-2\cdot\frac{\left(a+b+c\right)}{2}=3\)