VT=\(\frac{a^2}{ab+\frac{1}{b}}+\frac{b^2}{bc+\frac{1}{c}}+\frac{c^2}{ca+\frac{1}{a}}\)

áp dụng bđt cộng mẫu đc VT \(\ge\frac{\left(a+b+c\right)^2}{ab+bc+ca+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{\left(a+b+c\right)^2}{ab+bc+ca+\frac{ab+bc+ca}{abc}}\left(1\right)\)

Ta có  \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\forall a,b,c\)

Nên \(\left(1\right)\ge\frac{\left(a+b+c\right)^2}{\frac{\left(a+b+c\right)^2}{3}+\frac{\left(a+b+c\right)^2}{3abc}}=\frac{1}{\frac{1}{3}+\frac{1}{3abc}}=\frac{3abc}{1+abc}\left(đccm\right)\)

dấu bằng xảy ra <> a=b=c

\(\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}=\sqrt{\dfrac{ab+2c^2}{a^2+b^2+ab}}\)\(=\dfrac{ab+2c^2}{\sqrt{\left(a^2+b^2+ab\right)\left(ab+c^2+c^2\right)}}\)\(\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\)\(\ge\dfrac{2\left(ab+2c^2\right)}{2\left(a^2+b^2\right)+2c^2}\)\(=\dfrac{ab+2c^2}{a^2+b^2+c^2}\)

\(\Rightarrow\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}\ge ab+2c^2\)

Tương tự: \(\sqrt{\dfrac{bc+2a^2}{1+bc-a^2}}\ge bc+2a^2\); \(\sqrt{\dfrac{ac+2b^2}{1+ac-b^2}}\ge ac+2b^2\)

Cộng vế với vế \(\Rightarrow VT\ge2a^2+2b^2+2c^2+ab+bc+ac=2+ab+bc+ac\)

Dấu = xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)

cho a, b, c là các số thực dương thỏa mạn abc=1 chứng minh rằng a/(2b+a) +b/(2c+b)+c/(2a+c)>=1

Với x,y>0 ta cm: \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\Leftrightarrow\frac{x+y}{xy}\ge\frac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)

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

ÁP dụng vào bài toán ta có: 

\(\frac{1}{a+b+2c}=\frac{1}{a+c+b+c}\le\frac{1}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\)

\(\Rightarrow\frac{4ab}{a+b+2c}\le\frac{ab}{a+c}+\frac{ab}{b+c}\)

tương tự: \(\frac{4bc}{b+c+2a}\le\frac{bc}{a+b}+\frac{bc}{a+c};\frac{4ca}{c+a+2b}\le\frac{ca}{b+c}+\frac{ca}{a+b}\)

Cộng 3 bđt trên vế theo vế ta dc \(4\left(\frac{ab}{a+b+2c}+\frac{bc}{b+c+2a}+\frac{ca}{c+a+2b}\right)\le\frac{bc+ca}{a+b}+\frac{ab+ca}{b+c}+\frac{ab+bc}{a+c}=c+a+b\)

=>đpcm

Dấu "=" xảy ra <=> a=b=c

\(\frac{ab}{a+b+2c}=\frac{ab}{\left(a+c\right)+\left(b+c\right)}\le\frac{1}{4}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)\) 

Làm tương tự với 2 phân thức còn lại rồi cộng vào ra đpcm 

Bài này ta dùng bđt Cauchy-Schwaz

VT=\(\frac{\left(bc\right)^2}{a^2bc\left(b+c\right)}\)\(+\frac{\left(\text{c}\text{a}\right)^2}{\text{b}^2c\text{a}\left(\text{c}+\text{a}\right)}\)\(+\frac{\left(\text{a}\text{b}\right)^2}{\text{c}^2\text{a}\text{b}\left(\text{a}+b\right)}\)

\(\ge\)\(\frac{\left(ab+bc+ca\right)^2}{2abc\left(ab+bc+ca\right)}\)\(=\frac{ab+bc+ca}{2abc}\)\(=\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\)\(=\)VP

=> đpcm

Dấu \("="\)xảy ra <=> a=b=c