a,b,c>0 => ab+bc+ca=0 Amazing !!

Nhầm ab+ac+bc>=3

\(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)

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

\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)( phân tích nhân tử các kiểu )

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-abc\left(1\right)\)

\(a+b+c\ge3\sqrt[3]{abc};ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)

\(\Rightarrow-abc\ge\frac{-\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)

Khi đó:\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)

\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)

\(=\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\left(2\right)\)

Từ ( 1 ) và ( 2 ) có đpcm

Akai HarumaNguyễn Thành Trương

<= 3/4 nha ko phải a+b+c

minhf lm dc cau nay rồi

1. BĐT ban đầu

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

<=>\(\frac{a}{a+3b}+\frac{b}{b+3c}+\frac{c}{c+3a}\ge\frac{3}{4}\)

<=> \(\frac{a^2}{a^2+3ab}+\frac{b^2}{b^2+3bc}+\frac{c^2}{c^2+3ac}\ge\frac{3}{4}\)

Áp dụng BĐT buniacoxki dang phân thức 

=> BĐT cần CM

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

<=> \(a^2+b^2+c^2\ge ab+bc+ac\)luôn đúng 

=> BĐT được CM

2) \(a+b+c\le ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\)\(\Leftrightarrow\)\(\left(a+b+c\right)^2-3\left(a+b+c\right)\ge0\)

\(\Leftrightarrow\)\(\left(a+b+c\right)\left(a+b+c-3\right)\ge0\)\(\Leftrightarrow\)\(a+b+c\ge3\)

ko mất tính tổng quát giả sử \(a\ge b\ge c\)

Có: \(3\le a+b+c\le ab+bc+ca\le3a^2\)\(\Leftrightarrow\)\(3a^2\ge3\)\(\Leftrightarrow\)\(a\ge1\)

=> \(\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\le\frac{3}{1+2a}\le1\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=1\)

Bạn @Diệu Linh@ làm nhầm dòng 5 rồi nhé

2, BĐT ban đầu 

<=> \(\left(1-\frac{1}{1+a+b}\right)+\left(1-\frac{1}{1+b+c}\right)+\left(1-\frac{1}{1+a+c}\right)\ge2\)

<=> \(\frac{\left(a+b\right)^2}{a+b+\left(a+b\right)^2}+\frac{\left(b+c\right)^2}{b+c+\left(b+c\right)^2}+\frac{\left(c+a\right)^2}{c+a+\left(c+a\right)^2}\ge2\)

Dùng BĐT buniacoxki dạng phân thức ở VT 

\(VT\ge\frac{4\left(a+b+c\right)^2}{2\left(a+b+c\right)+\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2}\)

Mà \(a+b+c\le ab+bc+ac\)

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

Dấu bằng xảy ra khi a=b=c=1

 Đề bài bị trái dấu bạn nhé

CM \(\frac{5b^3-a^3}{ab+3b^2}\le2b-a\) 

\(\Leftrightarrow5b^3-a^3\le\left(2b-a\right)\left(ab+3b^2\right)\) 

\(\Leftrightarrow5b^3-a^3\le2ab^2+6b^3-a^2b-3ab^2\) 

\(\Leftrightarrow b^3+a^3-ab^2-ba^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\)đúng với mọi a, b>0 

CMTT các hạng tử khác 

\(\Rightarrow P=\frac{5b^3-a^3}{ab+3b^3}+\frac{5c^3-b^3}{bc+3c^3}+\frac{5a^3-c^3}{ac+3a^2}\le2b-a+2c-b+2a-c=a+b+c\)

vậy đề sai rồi chứ mình giải mãi chả ra mà toàn ngược dấu nên mình tưởng mình sai 

Áp dụng BĐT Bunhiacopxki ta có:

\(\left(1+1+1\right)\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

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

Dấu " = " xảy ra <=> a=b=c=1

Có: \(ab+bc+ca\le\frac{1}{3}\left(a+b+c\right)^2\Leftrightarrow a+b+c\ge3\)( bạn tự c/m nhé )

Dấu " = " xảy ra <=> a=b=c

Áp dụng BĐT Cauchy-schwarz ta có:

\(\frac{a^4}{b+3c}+\frac{b^4}{c+3a}+\frac{c^4}{a+3b}\ge\frac{\left(a^2+b^2+c^2\right)^2}{4\left(a+b+c\right)}\ge\frac{\left[\frac{\left(a+b+c\right)^2}{3}\right]^2}{4\left(a+b+c\right)}=\frac{\left(a+b+c\right)^3}{36}\ge\frac{27}{36}=\frac{3}{4}\)

Dấu " = " xảy ra <=> a=b=c=1 ( bạn tự giải rõ ra nhé )

Áp dụng bất đẳng thức Cauchy-Schwartz ta có

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

Tương tự ta có 2 bất đẳng thức khác nữa

\(\frac{bc}{b+3c+2a}=\frac{bc}{\left(b+a\right)+\left(a+c\right)+2c}\le\frac{bc}{9}\left(\frac{1}{b+a}+\frac{1}{a+c}+\frac{1}{2c}\right).\)

\(\frac{ac}{c+3a+2b}=\frac{ac}{\left(a+b\right)+\left(b+a\right)+2a}\le\frac{ac}{9}\left(\frac{1}{c+b}+\frac{1}{b+a}+\frac{1}{2a}\right).\)

Cộng ba bất đẳng thức lại cho ta \(\frac{ab}{a+3b+2c}+\frac{bc}{b+3c+2a}+\frac{ca}{c+3a+2b}\le\)

\(\le\frac{ab}{9}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{2b}\right)+\frac{bc}{9}\left(\frac{1}{b+a}+\frac{1}{a+c}+\frac{1}{2c}\right)+\frac{ac}{9}\left(\frac{1}{c+b}+\frac{1}{b+a}+\frac{1}{2a}\right)\)

\(=\frac{1}{9}\left(\frac{ab}{a+c}+\frac{bc}{a+c}\right)+\frac{1}{9}\left(\frac{ab}{b+c}+\frac{ac}{b+c}\right)+\frac{1}{9}\left(\frac{bc}{a+b}+\frac{ac}{a+b}\right)+\frac{a}{18}+\frac{b}{18}+\frac{c}{18}\)

\(=\frac{a+b+c}{6}.\)   (ĐPCM)

Áp dụng BĐT Bunhiacopxki ta có :

\(\left(1+1+1\right)\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

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

Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)

Ta có : \(ab+bc+ca\le\frac{1}{3}\left(a+b+c\right)^2\Leftrightarrow a+b+c\ge3\) ( tự chứng minh ạ )

Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)

Áp dụng BĐT Cachy Schwarz ta có :

\(\frac{a^4}{b+3c}+\frac{b^4}{c+3a}+\frac{c^4}{a+3b}\ge\frac{\left(a^2+b^2+c^2\right)^2}{4\left(a+b+c\right)}\) \(\ge\frac{\left[\frac{\left(a+b+c\right)}{3}\right]^2}{4\left(a+b+c\right)}=\frac{\left(a+b+c\right)^3}{36}\)

\(\ge\frac{27}{36}=\frac{3}{4}\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\) ( bạn tự giải rõ ạ )