Sử dụng bất đẳng thức Bunhiacopxki dạng phân thức và khi đó ta được:

\(\frac{a^5}{a^2+ab+b^2}+\frac{b^5}{b^2+bc+c^2}+\frac{c^5}{c^2+ca+a^2}\ge\)

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

\(\Rightarrow\)Ta cần chỉ ra được:

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

Hay: \(2\left(a^3+b^3+c^3\right)\ge a^2b+ab^2+b^2c+bc^2+c^2a+ca^2\)

Dễ thấy: \(a^3+b^3\ge ab\left(a+b\right);b^3+c^3\ge bc\left(b+c\right);c^3+a^3\ge ca\left(c+a\right)\)

Cộng theo vế các bất đẳng thức trên ta được:

\(2\left(a^3+b^3+c^3\right)\ge a^2b+ab^2+b^2c+bc^2+c^2a+ca^2\)

Vậy bất đẳng thức đã được chứng minh.

chứng minh \(\sqrt{2x+1}\)là số vô tỉ

1.

Áp dụng BĐT \(x^2+y^2+z^2\ge xy+yz+zx\)

\(\Rightarrow\left(\sqrt{ab}\right)^2+\left(\sqrt{bc}\right)^2+\left(\sqrt{ca}\right)^2\ge\sqrt{ab}.\sqrt{bc}+\sqrt{ab}.\sqrt{ac}+\sqrt{bc}.\sqrt{ac}\)

\(\Rightarrow ab+bc+ca\ge\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)

2.

\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt[]{\frac{ab.bc}{ca}}=2b\) ; \(\frac{ab}{c}+\frac{ac}{b}\ge2a\) ; \(\frac{bc}{a}+\frac{ac}{b}\ge2c\)

Cộng vế với vế:

\(2\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\right)\ge2\left(a+b+c\right)\)

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

3.

Từ câu b, thay \(c=1\) ta được:

\(ab+\frac{b}{a}+\frac{a}{b}\ge a+b+1\)

4.

\(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}=\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ac}\ge\frac{\left(a^2+b^2+c^2\right)}{ab+bc+ca}\)

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

Dấu "=" xảy ra khi \(a=b=c\)

5.

\(\frac{a}{bc}+\frac{b}{ca}\ge2\sqrt{\frac{ab}{bc.ca}}=\frac{2}{c}\) ; \(\frac{a}{bc}+\frac{c}{ab}\ge\frac{2}{b}\) ; \(\frac{b}{ca}+\frac{c}{ab}\ge\frac{2}{a}\)

Cộng vế với vế:

\(2\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\Rightarrow\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

1. bđt được viết lại thành

\(ab+bc+ca\ge a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\)

Theo bđt AM-GM thì :

\(ab+bc\ge2\sqrt{ab\cdot bc}=2\sqrt{ab^2c}=2b\sqrt{ac}\)

Tương tự : \(bc+ca\ge2c\sqrt{ab}\); \(ab+ca\ge2a\sqrt{bc}\)

Cộng vế với vế

=> \(2\left(ab+bc+ca\right)\ge2\left(a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\right)\)

=> \(ab+bc+ca\ge a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\)( đpcm )

Dấu "=" xảy ra <=> a=b=c

Áp dụng BĐT Cauchy Shwarz dạng Engel và BĐT AM - GM, ta có:

\(\frac{a^5}{bc}+\frac{b^5}{ac}+\frac{c^5}{ab}\)

\(=\frac{a^6}{abc}+\frac{b^6}{abc}+\frac{c^6}{abc}\)

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

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

\(=a^3+b^3+c^3\left(\text{đ}pcm\right)\)

Dấu "=" xảy ra khi a = b = c

\(\frac{a^5}{bc}+\frac{b^5}{ca}+\frac{c^5}{ab}=\frac{1}{abc}\left(a^6+b^6+c^6\right)\)

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

Áp dụng BĐT AM-GM ta có: \(\frac{a}{b^3+ab}=\frac{1}{b}-\frac{b}{a+b^2}\ge\frac{1}{b}-\frac{b}{2\sqrt{ab^2}}=\frac{1}{b}-\frac{1}{2\sqrt{a}}\ge\frac{1}{b}-\frac{1}{4}\left(\frac{1}{a}+1\right)\)

Tương tự có: \(\hept{\begin{cases}\frac{b}{c^3+ca}\ge\frac{1}{c}-\frac{1}{4}\left(\frac{1}{b}+1\right)\\\frac{c}{a^3+ca}\ge\frac{1}{a}-\frac{1}{4}\left(\frac{1}{c}+1\right)\end{cases}}\)

Cộng 3 vế BĐT ta được:  \(\frac{a}{b^3+ab}+\frac{b}{c^3+bc}+\frac{c}{a^3+ca}\ge\frac{3}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{3}{4}\)

Bài toán quy về chứng minh \(\frac{3}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{3}{4}\ge\frac{3}{2}\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\Leftrightarrow\left(\frac{1}{a}+a\right)\left(\frac{1}{b}+b\right)\left(\frac{1}{c}+c\right)\ge3+a+b+c=6\)

BĐT cuối hiển nhiên đúng vì theo BĐT AM-GM ta có:

\(\hept{\begin{cases}\frac{1}{a}+a\ge2\\\frac{1}{b}+b\ge2\\\frac{1}{c}+c\ge2\end{cases}}\)

Dấu "=" xảy ra <=> a=b=c=1

\(\frac{a}{b^3+ab}+\frac{b}{c^3+bc}+\frac{c}{a^3+ac}\)

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

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

\(\ge\frac{1}{b}-\frac{b}{2b\sqrt{a}}+\frac{1}{c}-\frac{c}{2c\sqrt{b}}+\frac{1}{a}-\frac{a}{2a\sqrt{c}}\)

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

\(\ge\frac{3}{4}.\frac{9}{a+b+c}+\frac{1}{4}\left(\frac{1}{\sqrt{a}}-1\right)^2+\frac{1}{4}\left(\frac{1}{\sqrt{b}}-1\right)^2+\frac{1}{4}\left(\frac{1}{\sqrt{b}}-1\right)^2-\frac{3}{4}\)

\(\ge\frac{3}{2}\)

Dấu "=" xảy ra <=> a = b = c = 1.

1,

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

a) Áp dụng bất đẳng thức AM-GM ta có ngay :

\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt{\frac{ab}{c}\cdot\frac{bc}{a}}=2\sqrt{\frac{ab^2c}{ac}}=2\sqrt{b^2}=2\left|b\right|=2b\)( do b > 0 )

=> đpcm

Đẳng thức xảy ra <=> a = b = c

b) Áp dụng bất đẳng thức AM-GM ta có :

\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt{\frac{ab}{c}\cdot\frac{bc}{a}}=2b\)(1) ( như a) đấy :)) )

tương tự : \(\frac{bc}{a}+\frac{ca}{b}\ge2c\)(2) ; \(\frac{ab}{c}+\frac{ca}{b}\ge2a\)(3)

Cộng (1), (2), (3) theo vế ta có đpcm

Đẳng thức xảy ra <=> a = b = c

c) \(\frac{a^3+b^3}{2ab}+\frac{b^3+c^3}{2bc}+\frac{c^3+a^3}{2ca}\)

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

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

Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

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

hay \(\frac{a^3+b^3}{2ab}+\frac{b^3+c^3}{2bc}+\frac{c^3+a^3}{2ca}\ge a+b+c\)(đpcm)

Đẳng thức xảy ra <=> a = b = c