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

\(\Leftrightarrow\left(c^2b-abc-b^2c+ab^2\right)+\left(ca^2+abc-ac^2-a^2b\right)\ge0\)

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

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

\(\Leftrightarrow\left(b-a\right)\left(c-b\right)\left(c-a\right)\ge0\) (luôn đúng do \(c\ge b\ge a>0\))

a.

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

2.

\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{a+b+2c+2b}\le\dfrac{ab}{9}\left(\dfrac{4}{a+b+2c}+\dfrac{1}{2b}\right)=4.\dfrac{ab}{a+b+2c}+\dfrac{a}{18}\)

Quay lại câu a

\(b,\dfrac{ab}{a+3b+2c}=\left(\dfrac{1}{9}ab\right)\cdot\dfrac{9}{\left(a+c\right)+\left(b+c\right)+2b}\le\left(\dfrac{1}{9}ab\right)\cdot\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{2b}\right)=\dfrac{1}{9}\cdot\left(\dfrac{ab}{a+b}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)

Cmtt: \(\dfrac{bc}{b+3c+2a}\le\dfrac{1}{9}\cdot\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+b}+\dfrac{b}{2}\right);\dfrac{ca}{c+3a+2b}\le\dfrac{1}{9}\cdot\left(\dfrac{ca}{b+c}+\dfrac{ca}{a+b}+\dfrac{c}{2}\right)\)

\(\Rightarrow VT\le\dfrac{1}{9}\left(\dfrac{bc+ca}{a+b}+\dfrac{ab+ac}{b+c}+\dfrac{ab+bc}{a+c}+\dfrac{a+b+c}{2}\right)\\ \le\dfrac{1}{9}\left(a+b+c+\dfrac{a+b+c}{2}\right)=\dfrac{1}{9}\cdot\dfrac{3}{2}\left(a+b+c\right)=\dfrac{a+b+c}{6}\)

Dấu $"="$ khi $a=b=c$

Lời giải:
Xét hiệu:

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

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

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

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

\(=\frac{(b-a)(b-c)}{bc}+\frac{(c-b)(b-a)}{ab}=(b-a)(b-c)(\frac{1}{bc}-\frac{1}{ab})\)

\(=\frac{(b-a)(b-c)(a-c)}{abc}\geq 0\) do \(0\leq a\leq b\leq c\)

Do đó:
\(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{b}{a}+\frac{c}{b}+\frac{a}{c}\) (đpcm)

Dấu bằng xảy ra khi $a=b=c$

3/ Áp dụng bất đẳng thức AM-GM, ta có :

\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)

\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)

\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)

Cộng 3 vế của BĐT trên ta có :

\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)

\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)

Bài 1:

Áp dụng BĐT AM-GM ta có:

\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)

Tiếp tục áp dụng BĐT AM-GM:

\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)

Do đó:

\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)

Dấu "=" xảy ra khi $a=b=c$

Bài 2:

Thay $1=a+b+c$ và áp dụng BĐT AM-GM ta có:

\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{(a+1)(b+1)(c+1)}{abc}\)

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

\(\geq \frac{4\sqrt[4]{a.a.b.c}.4\sqrt[4]{b.a.b.c}.4\sqrt[4]{c.a.b.c}}{abc}=\frac{64abc}{abc}=64\)

Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$

a) Áp dụng bất đẳng thức AM-GM ta có:

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

Chứng minh tương tự ta được: \(\left\{{}\begin{matrix}\dfrac{ac}{b}+\dfrac{ab}{c}\ge2a\\\dfrac{bc}{a}+\dfrac{ab}{c}\ge2b\end{matrix}\right.\)

Cộng theo vế: \(\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\ge a+b+c\left(đpcm\right)\)

Áp dụng liên tiếp AM-GM và Cauchy-Schwarz ta được:

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

Chứng minh tương tự:

\(\left\{{}\begin{matrix}\dfrac{bc}{b+c}=\dfrac{bc+c^2-c^2}{b+c}=\dfrac{c\left(b+c\right)}{b+c}-\dfrac{c^2}{b+c}=c-\dfrac{c^2}{b+c}\\\dfrac{ac}{c+a}=\dfrac{ac+a^2-a^2}{c+a}=\dfrac{a\left(c+a\right)}{c+a}-\dfrac{a^2}{c+a}=a-\dfrac{a^2}{c+a}\end{matrix}\right.\)

Cộng theo vế:

\(\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ac}{a+c}=a+b+c-\left(\dfrac{b^2}{a+b}+\dfrac{c^2}{b+c}+\dfrac{a^2}{a+c}\right)\le\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=a+b+c-\dfrac{a+b+c}{2}=\dfrac{a+b+c}{2}\left(đpcm\right)\)

b)Đặt \(A=\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ca}{c+a}\)

\(A=\dfrac{a\left(a+b\right)-a^2}{a+b}+\dfrac{b\left(b+c\right)-b^2}{a+b}+\dfrac{c\left(c+a\right)-c^2}{c+a}\)

\(A=a+b+c-\dfrac{a^2}{a+b}-\dfrac{b^2}{b+c}-\dfrac{c^2}{c+a}\)

Lại có:\(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)

\(\Rightarrow A\le a+b+c-\dfrac{a+b+c}{2}=\dfrac{a+b+c}{2}\)

\(\Rightarrowđpcm\)

Bài 3:

\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\)

\(\Leftrightarrow x^2y^2\left(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\ge\dfrac{4}{xy}.x^2y^2\)

\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2+y^2\ge4xy\)

\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2-2xy+y^2\ge2xy\)

\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2+\left(x-y\right)^2\ge2xy\)

\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2-2xy+\left(x-y\right)^2\ge0\)

\(\Leftrightarrow\left(\dfrac{xy}{x-y}-x+y\right)^2=0\) (luôn đúng)

-Tham khảo:

-Tham khảo:

Áp dụng bđt cosi ta có:

`a+b>=2sqrt{ab}`

`=>(ab)/(a+b)<=(sqrt{ab})/2`

Chứng minh tt:

`(bc)/(b+c)<=(sqrt{bc})/2`

`(ca)/(a+c)<=(sqrt{ca})/2`

`=>VT<=(sqrt{ab}+sqrt{bc}+sqrt{ca})/2`

Áp dụng cosi:

`sqrt{ab}<=(a+b)/2`

`sqrt{bc}<=(b+c)/2`

`sqrt{ca}<=(c+a)/2`

`=>(sqrt{ab}+sqrt{bc}+sqrt{ca})/2<=(a+b+c)/2`

`=>VT<=(a+b+c)/2`

\(\dfrac{a+b}{ab+c^2}=\dfrac{\left(a+b\right)^2}{\left(ab+c^2\right)\left(a+b\right)}=\dfrac{\left(a+b\right)^2}{b\left(a^2+c^2\right)+a\left(b^2+c^2\right)}\le\dfrac{a^2}{b\left(a^2+c^2\right)}+\dfrac{b^2}{a\left(b^2+c^2\right)}\)

Tương tự: 

\(\dfrac{b+c}{bc+a^2}\le\dfrac{b^2}{c\left(a^2+b^2\right)}+\dfrac{c^2}{b\left(a^2+c^2\right)}\) ; \(\dfrac{c+a}{ca+b^2}\le\dfrac{c^2}{a\left(b^2+c^2\right)}+\dfrac{a^2}{c\left(a^2+b^2\right)}\)

Cộng vế:

\(VT\le\dfrac{1}{a}\left(\dfrac{b^2}{b^2+c^2}+\dfrac{c^2}{b^2+c^2}\right)+\dfrac{1}{b}\left(\dfrac{a^2}{a^2+c^2}+\dfrac{c^2}{a^2+c^2}\right)+\dfrac{1}{c}\left(\dfrac{a^2}{a^2+b^2}+\dfrac{b^2}{a^2+b^2}\right)=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)