\(\frac{a^3}{b^3}+1+1\ge\frac{3a}{b}\) ; \(\frac{b^3}{c^3}+1+1\ge\frac{3b}{c}\) ; \(\frac{c^3}{a^3}+1+1\ge\frac{3c}{a}\)

Cộng vế với vế:

\(\frac{a^3}{b^3}+\frac{b^3}{c^3}+\frac{c^3}{a^3}+6\ge\frac{3a}{b}+\frac{3b}{c}+\frac{3c}{a}\)

\(\Leftrightarrow\frac{a^3}{b^3}+\frac{b^3}{c^3}+\frac{c^3}{a^3}\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-6\)

\(\Rightarrow\frac{a^3}{b^3}+\frac{b^3}{c^3}+\frac{c^3}{a^3}\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+2.3\sqrt[3]{\frac{abc}{bca}}-6=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

Bài 1 :

Áp dụng BĐT Cô - si cho 3 số không âm

\(\sqrt{\frac{a^3}{b^3}}+\sqrt{\frac{a^3}{b^3}}+1\ge3\sqrt[3]{\sqrt{\frac{a^6}{b^6}}}=\frac{3a}{b}\)

\(\sqrt{\frac{b^3}{c^3}}+\sqrt{\frac{b^3}{c^3}}+1\ge3\sqrt[3]{\sqrt{\frac{b^6}{c^6}}}=\frac{3b}{c}\)

\(\sqrt{\frac{c^3}{a^3}}+\sqrt{\frac{c^3}{a^3}}+1\ge3\sqrt[3]{\sqrt{\frac{c^6}{a^6}}}=\frac{3c}{a}\)

Cộng theo vế , ta được :

\(2\left(\sqrt{\frac{a^3}{b^3}}+\sqrt{\frac{b^3}{c^3}}+\sqrt{\frac{c^3}{a^3}}\right)+3\ge2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\)

\(\ge2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)+3\)

\(\Rightarrow2\left(\sqrt{\frac{a^3}{b^3}}+\sqrt{\frac{b^3}{c^3}}+\sqrt{\frac{c^3}{a^3}}\right)\ge2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)

\(\Rightarrow\left(\sqrt{\frac{a^3}{b^3}}+\sqrt{\frac{b^3}{c^3}}+\sqrt{\frac{c^3}{a^3}}\right)\ge\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)

Vậy \(\Rightarrow\left(\sqrt{\frac{a^3}{b^3}}+\sqrt{\frac{b^3}{c^3}}+\sqrt{\frac{c^3}{a^3}}\right)\ge\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\left(đpcm\right)\)

giả sử a\(\le\)b \(\le\)c.

khi đó \(\frac{a}{b+c}\le\frac{b}{c+a}\le\frac{c}{a+b}\)

áp dụng BĐT Trê bư sép ta có:

\(\left(a^2+b^2+c^2\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\le3\left(\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}\right)=3VT\)

lại có a² + b² + c² \(\ge\) \(\frac{\left(a+b+c\right)^2}{3}\) nên:

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

hay VT \(\ge\left(\frac{a+b+c}{3}\right)^2\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\). đpcm

Áp dụng BĐT Cô - si cho 3 số không âm:

\(\sqrt{\frac{a^3}{b^3}}+\sqrt{\frac{a^3}{b^3}}+1\ge3\sqrt[3]{\sqrt{\frac{a^6}{b^6}}}=\frac{3a}{b}\)

\(\sqrt{\frac{b^3}{c^3}}+\sqrt{\frac{b^3}{c^3}}+1\ge3\sqrt[3]{\sqrt{\frac{b^6}{c^6}}}=\frac{3b}{c}\)

\(\sqrt{\frac{c^3}{a^3}}+\sqrt{\frac{c^3}{a^3}}+1\ge3\sqrt[3]{\sqrt{\frac{c^6}{a^6}}}=\frac{3c}{a}\)

Cộng vế theo vế ,ta được:

\(2\left(\sqrt{\frac{a^3}{b^3}}+\sqrt{\frac{b^3}{c^3}}+\sqrt{\frac{c^3}{a^3}}\right)+3\ge2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)\(+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\)

\(\ge2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)\(+3\)

\(\Rightarrow2\left(\sqrt{\frac{a^3}{b^3}}+\sqrt{\frac{b^3}{c^3}}+\sqrt{\frac{c^3}{a^3}}\right)\ge2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)

\(\Rightarrow\left(\sqrt{\frac{a^3}{b^3}}+\sqrt{\frac{b^3}{c^3}}+\sqrt{\frac{c^3}{a^3}}\right)\ge\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)

Vậy \(\Rightarrow\left(\sqrt{\frac{a^3}{b^3}}+\sqrt{\frac{b^3}{c^3}}+\sqrt{\frac{c^3}{a^3}}\right)\ge\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)(đpcm)

Trâu bò chút!

Đặt \(\sqrt{\frac{a}{b}}=x;\sqrt{\frac{b}{c}}=y;\sqrt{\frac{c}{a}}=z\Rightarrow xyz=1\)

BĐT quy về chứng minh: \(x^3+y^3+z^3\ge x^2+y^2+z^2\)

Để ý rằng: \(x^3=\frac{\left(x-1\right)^2\left(2x+1\right)}{2}+\frac{3}{2}x^2-\frac{1}{2}\)

Từ đó ta có:  \(VT-VP=\Sigma_{cyc}\frac{\left(x-1\right)^2\left(2x+1\right)}{2}+\frac{1}{2}\left(\Sigma x^2-3\right)\)

\(\ge\Sigma_{cyc}\frac{\left(x-1\right)^2\left(2x+1\right)}{2}\ge0\)

P/s: Nếu thích troll người thì thế ngược lại các biến đã đặt ta tìm được:

\(VT-VP\ge\Sigma_{cyc}\frac{\left(a-b\right)^2\left(2\sqrt{a}+\sqrt{b}\right)}{2b\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)^2}\ge0\)

 a + b + c = 3 

a/( 1 + b^2 ) + b/( 1 + c^2 ) + c/( 1 + a^2 ) ≥ 3/2 

Ta có 

a/( 1 + b^2 ) = a - ab^2/( 1 + b^2 ) ≥ a - ab^2/2b = a - ab/2 

Tương tự ta có 

b/( 1 + c^2 ) ≥ b - bc/2 

c/( 1 + a^2 ) ≥ c - ac/2 

Cộng vào ta có 

a/( 1 + b^2 ) + b/( 1 + c^2 ) + c/( 1 + a^2 ) ≥ a + b + c - ( ab + bc + ac )/2 = 3 - ( ab + bc + ac )/2 

Xét ab + bc + ac 

Ta có 

a^2 + b^2 ≥ 2ab 
b^2 + c^2 ≥ 2bc 
c^2 + a^2 ≥ 2ac 

=> a^2 + b^2 + c^2 ≥ ab + bc + ac 

<=> a^2 + b^2 + c^2 + 2ac + 2bc + 2ab ≥ 3( ab + ac + bc ) 

<=> ( a + b + c )^2 ≥ 3( ab + ac + bc ) 

<=> ab + ac + bc ≤ 9:3 = 3 

=> 3 - ( ab + bc + ac )/2 ≥ 3 - 3/2 = 3/2 

=> a/( 1 + b^2 ) + b/( 1 + c^2 ) + c/( 1 + a^2 ) ≥   

Theo dãy tỉ số bằng nhau , có :

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

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

\(\Rightarrow\left(\frac{b+c}{a}\right)^3=\left(\frac{c+a}{b}\right)^3=\left(\frac{a+b}{c}\right)^3=8\)

\(\Rightarrow\frac{\left(b+c\right)^3}{a^3}+\frac{\left(c+a\right)^3}{b^3}+\frac{\left(a+b\right)^3}{c^3}=8.3=24\)

a/ BĐT sai, cho \(a=b=c=2\) là thấy

b/ \(VT=\frac{a^4}{a^2+2ab}+\frac{b^4}{b^2+2bc}+\frac{c^4}{c^2+2ac}\ge\frac{\left(a^2+b^2+c^2\right)^2}{\left(a+b+c\right)^2}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)}{\left(a+b+c\right)^2}\)

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

Dấu "=" xảy ra khi \(a=b=c\)

c/ Tiếp tục sai nữa, vế phải là \(\frac{3}{2}\) chứ ko phải \(2\), và hy vọng rằng a;b;c dương

\(VT=\frac{a^2}{abc.b+a}+\frac{b^2}{abc.c+b}+\frac{c^2}{abc.a+c}\ge\frac{\left(a+b+c\right)^2}{abc\left(a+b+c\right)+a+b+c}\)

\(VT\ge\frac{9}{3abc+3}\ge\frac{9}{\frac{3\left(a+b+c\right)^3}{27}+3}=\frac{9}{\frac{3.3^3}{27}+3}=\frac{9}{6}=\frac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Ta có:

\(a^3+b^3+b^3\ge3ab^2\) ; \(b^3+c^3+c^3\ge3bc^2\) ; \(c^3+a^3+a^3\ge3ca^2\)

Cộng vế với vế \(\Rightarrow a^3+b^3+c^3\ge ab^2+bc^2+ca^2\)

\(\frac{a^5}{b^2}+\frac{b^5}{c^2}+\frac{c^5}{a^2}=\frac{a^6}{ab^2}+\frac{b^6}{bc^2}+\frac{c^6}{ca^2}\ge\frac{\left(a^3+b^3+c^3\right)^2}{ab^2+bc^2+ca^2}\ge\frac{\left(a^3+b^3+c^3\right)^2}{a^3+b^3+c^3}=a^3+b^3+c^3\)