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

Đề chơi căng nhỉ?

a) Dễ chứng minh VP =< 3

BĐT \(\Leftrightarrow\left(\frac{a+b}{1+a}-1\right)+\left(\frac{b+c}{1+b}-1\right)+\left(\frac{c+a}{1+c}-1\right)\ge0\)

\(\Leftrightarrow\frac{b-1}{1+a}+\frac{c-1}{1+b}+\frac{a-1}{1+c}\ge0\)

\(\Leftrightarrow\frac{\left(b-1\right)^2}{\left(1+a\right)\left(b-1\right)}+\frac{\left(c-1\right)^2}{\left(1+b\right)\left(c-1\right)}+\frac{\left(a-1\right)^2}{\left(1+c\right)\left(a-1\right)}\) >=0

Áp dụng BĐT Cauchy-Schwarz dạng Engel vào VT ta có đpcm.

P/s: Èo, sao đơn giản thế nhỉ? Em có làm sai chỗ nào chăng?

a, Ta có \(\frac{a+b}{a+1}=\frac{\left(a+b\right)\left(a+1\right)-a\left(a+b\right)}{a+1}=a+b-\frac{a\left(a+b\right)}{a+1}\)

Mà \(\frac{1}{a+1}\le\frac{a+1}{4a}\)

=> \(\frac{a+b}{1+a}\ge a+b-\frac{\left(a+1\right)\left(a+b\right)}{4}=\frac{3}{4}\left(a+b+c\right)-\frac{1}{4}a^2-\frac{1}{4}ab\)

Khi đó

\(Vt\ge\frac{3}{2}\left(a+b+c\right)-\frac{1}{4}\left(a^2+b^2+c^2\right)-\frac{1}{4}\left(ab+bc+ac\right)\)

=> \(VT\ge\frac{9}{2}-\frac{1}{4}\left(9-2ab-2bc-2ac\right)-\frac{1}{4}\left(ab+bc+ac\right)\)

=> \(VT\ge\frac{9}{4}+\frac{1}{4}\left(ab+bc+ac\right)\)

Lại có \(ab+bc+ac\le\frac{1}{3}\left(a+b+c\right)^2=3\)

=> \(VT\ge ab+bc+ac\)(ĐPCM)

Dấu bằng xảy ra khi a=b=c=1

b,Ta có \(\frac{a}{b\left(a+b^2\right)}=\frac{a+b^2-b^2}{b\left(a+b^2\right)}=\frac{1}{b}-\frac{b}{a+b^2}\)

Mà \(a+b^2\ge2b\sqrt{a}\)

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

Lại có \(\frac{1}{\sqrt{a.1}}\le\frac{1}{2}\left(\frac{1}{a}+1\right)\)

=> \(\frac{a}{b\left(a+b^2\right)}\ge\frac{1}{b}-\frac{1}{4}.\left(\frac{1}{a}+1\right)\)

Khi đó

\(VT\ge\frac{3}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{3}{4}\)

Mà \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}=3\)

=> \(VT\ge\frac{9}{4}-\frac{3}{4}=\frac{3}{2}\)(ĐPCM)

Dấu bằng xảy ra khi a=b=c=1

Bất đẳng thức được viết lại thành

\(\sum\frac{3-a}{1+a}\ge ab+bc+ca\)

Mà \(ab+bc+ca\le3\) nên ta chỉ cần chứng minh

\(\sum\frac{3-a}{1+a}\ge3\)

Ta chứng minh bất đẳng thức phụ sau

\(\frac{3-a}{1+a}\ge2-a\)

\(\Leftrightarrow\left(a-1\right)^2\ge0\)

Thiết lập các bất đẳng thức tương tự ta có điều phải chứng minh

\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\Leftrightarrow\frac{2+a^2+b^2}{\left(1+a^2+b^2+a^2b^2\right)}\ge\frac{2}{1+ab}\)

\(\Leftrightarrow\left(1+ab\right)\left(2+a^2+b^2\right)\ge2a^2b^2+2a^2+2b^2+2\)

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

\(\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2\ge0\)

b/ \(\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{2}{1+b^4}\ge\frac{2}{1+a^2b^2}+\frac{2}{1+b^4}\ge\frac{4}{1+ab^3}\)

\(\Rightarrow\frac{1}{1+a^4}+\frac{3}{1+b^4}\ge\frac{4}{1+ab^3}\)

Hoàn toàn tương tự: \(\frac{1}{1+b^4}+\frac{3}{1+c^4}\ge\frac{4}{1+bc^3}\); \(\frac{1}{1+c^4}+\frac{3}{1+a^4}\ge\frac{4}{1+a^3c}\)

Cộng vế với vế ta có đpcm

Lời giải:

Ta thấy:

\(\text{VT}=(a+\frac{ca}{a+b})+(b+\frac{ab}{b+c})+(c+\frac{bc}{c+a})\)

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

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

\(\geq (a+b+c).\frac{(a+b+c)^2}{a^2+ab+b^2+bc+c^2+ac}=\frac{(a+b+c)^3}{a^2+b^2+c^2+ab+bc+ac}\) (theo BĐT Cauchy-Schwarz)

Có:

$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)=a^2+b^2+c^2+2$

$\Rightarrow a+b+c=\sqrt{a^2+b^2+c^2+2}=\sqrt{t+2}$ với $t=a^2+b^2+c^2$

Do đó:

$\text{VT}\geq \frac{\sqrt{(t+2)^3}}{t+1}$ \(=\sqrt{\frac{(t+2)^3}{(t+1)^2}}\)

Áp dụng BĐT AM-GM:

\((t+2)^3=\left(\frac{t+1}{2}+\frac{t+1}{2}+1\right)^3\geq 27.\frac{(t+1)^2}{4}\)

\(\Rightarrow \text{VT}=\sqrt{\frac{(t+2)^3}{(t+1)^2}}\geq \sqrt{\frac{27}{4}}=\frac{3\sqrt{3}}{2}\) (đpcm)

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

Quay lại diễn đàn trong thinh lặng:))

Chứng minh: $$\left( a+{\frac {ab}{b+c}}+b+{\frac {bc}{c+a}}+c+{\frac {ca}{a+b}}
\right) ^{2}-{\frac {27\,ab}{4}}-{\frac {27\,ca}{4}} \geqq {\frac {27\,bc}{
4}}$$

Sau khi quy đồng, cần chứng minh$:$

$$\frac{1}{2} \sum\limits_{cyc} \left( 5\,{a}^{4}{b}^{2}+8\,{a}^{3}{b}^{3}+7\,{a}^{2}{b}^{4}+98\,{a}^
{2}{b}^{3}c+99\,{a}^{2}{b}^{2}{c}^{2}+124\,{a}^{2}b{c}^{3}+34\,a{b}^{4
}c+130\,a{b}^{3}{c}^{2}+26\,{b}^{4}{c}^{2}+44\,{b}^{3}{c}^{3}+{c}^{6}
\right) \left( a-b \right) ^{2} \geqq 0$$

Áp dụng BĐT AM-GM ta có:

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

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

\(\ge\dfrac{9}{a+b+c}-\left(\dfrac{a}{2a\sqrt{c}}+\dfrac{b}{2b\sqrt{a}}+\dfrac{c}{2c\sqrt{b}}\right)\)

\(\ge3-\left(\dfrac{1}{2\sqrt{c}}+\dfrac{1}{2\sqrt{a}}+\dfrac{1}{2\sqrt{b}}\right)\)\(=3-\left(\dfrac{2\sqrt{a}}{4a}+\dfrac{2\sqrt{b}}{4b}+\dfrac{2\sqrt{c}}{4c}\right)\)

\(\ge3-\left(\dfrac{a+1}{4a}+\dfrac{b+1}{4b}+\dfrac{c+1}{4c}\right)\)

\(=3-\left(\dfrac{3}{4}+\dfrac{1}{4a}+\dfrac{1}{4b}+\dfrac{1}{4c}\right)\ge3-\left(\dfrac{3}{4}+\dfrac{9}{4\left(a+b+c\right)}\right)=\dfrac{3}{2}\)

Khi \(a=b=c=1\)

\(\frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{ca+1}\ge\frac{9}{3+ab+ca+bc}\)

Cần c/m \(\frac{9}{3+ab+bc+ca}\ge\frac{9}{6}\Leftrightarrow ab+cb+ca\le3\)(*)

Mà \(a^2+b^2+c^2\ge ab+bc+ac\Rightarrow\left(a+b+c\right)^2\ge3ab+3ac+3bc\)

Mặt khác a+b+c=3

nên BĐT (*) đúng hay BĐT cần c/m luôn đúng

giúp với 

Sử dụng BĐT quen thuộc: \(\frac{1}{1+x^2}+\frac{1}{1+y^2}\ge\frac{2}{1+xy}\) với \(xy\ge1\)

\(2VT\ge\frac{2}{1+a^2b^2}+\frac{2}{1+b^2c^2}+\frac{2}{1+c^2a^2}\)

\(\Rightarrow VT\ge\frac{1}{1+a^2b^2}+\frac{1}{1+b^2c^2}+\frac{1}{1+c^2a^2}\)

\(\Rightarrow2VT\ge\frac{1}{1+a^2b^2}+\frac{1}{1+b^4}+\frac{1}{1+b^2c^2}+\frac{1}{1+c^4}\frac{1}{1+c^2a^2}+\frac{1}{1+a^4}\)

\(\Rightarrow2VT\ge\frac{2}{1+ab^3}+\frac{2}{1+bc^3}+\frac{2}{1+ca^3}\)

\(\Rightarrow VT\ge\frac{1}{1+ab^3}+\frac{1}{1+bc^3}+\frac{1}{1+ca^3}\)

Dấu "=" xảy ra khi \(a=b=c=1\)