Xét: \(\frac{a+3}{3a+bc}+\frac{b+3}{3b+ca}+\frac{c+3}{3c+ab}\)

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

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

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

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

Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm

\(\Rightarrow\left\{\begin{matrix}\left(a+b\right)\left(a+c\right)\le\left(\frac{2a+b+c}{2}\right)^2=\frac{\left(2a+b+c\right)^2}{4}\\\left(b+a\right)\left(b+c\right)\le\left(\frac{a+2b+c}{2}\right)^2=\frac{\left(a+2b+c\right)^2}{4}\\\left(a+c\right)\left(b+c\right)\le\left(\frac{a+b+2c}{2}\right)^2=\frac{\left(a+b+2c\right)^2}{4}\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}\frac{2a+b+c}{\left(a+b\right)\left(a+c\right)}\ge\frac{4\left(2a+b+c\right)}{\left(2a+b+c\right)^2}=\frac{4}{2a+b+c}\\\frac{a+2b+c}{\left(b+a\right)\left(b+c\right)}\ge\frac{4\left(a+2b+c\right)}{\left(a+2b+c\right)^2}=\frac{4}{a+2b+c}\\\frac{a+b+2c}{\left(a+c\right)\left(b+c\right)}\ge\frac{4\left(a+b+2c\right)}{\left(a+b+2c\right)^2}=\frac{4}{a+b+2c}\end{matrix}\right.\)

\(\Rightarrow VT\ge\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\)

Xét: \(\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\)

Áp dụng bất đẳng thức cộng mẫu số

\(\Rightarrow\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\ge\frac{\left(2+2+2\right)^2}{2a+b+c+a+2b+c+a+b+2c}=\frac{36}{4\left(a+b+c\right)}=\frac{36}{12}=3\)

Mà \(VT\ge\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\)

\(\Rightarrow VT\ge3\)

\(\Leftrightarrow\frac{a+3}{3a+bc}+\frac{b+3}{3b+ca}+\frac{c+3}{3c+ab}\ge3\) ( đpcm )

Ta có:

\(3a+bc=(a+b+c)a+bc=(a+c)(a+b)\)

\(\Rightarrow \sum \frac{a+3}{3a+bc}\)\(= \sum \frac{(a+c)+(a+b)}{(a+c)(a+b)}=2 \sum \frac{1}{a+b}\geq 2.\frac{9}{2(a+b+c)}=3\)

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

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

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

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

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

Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm 

\(\Rightarrow\hept{\begin{cases}\left(a+b\right)\left(a+c\right)\le\left(\frac{2a+b+c}{2}\right)^2=\frac{\left(2a+b+c\right)^2}{4}\\\left(b+a\right)\left(b+c\right)\le\left(\frac{a+2b+c}{2}\right)^2=\frac{\left(a+2b+c\right)^2}{4}\\\left(a+c\right)\left(b+c\right)\le\left(\frac{a+b+2c}{2}\right)^2=\frac{\left(a+b+2c\right)^2}{4}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\frac{2a+b+c}{\left(a+b\right)\left(a+c\right)}\ge\frac{4\left(2a+b+c\right)}{\left(2a+b+c\right)^2}=\frac{4}{2a+b+c}\\\frac{a+2b+c}{\left(b+a\right)\left(b+c\right)}\ge\frac{4\left(a+2b+c\right)}{\left(a+2b+c\right)^2}=\frac{4}{a+2b+c}\\\frac{a+b+2c}{\left(a+c\right)\left(b+c\right)}\ge\frac{4\left(a+b+2c\right)}{\left(a+b+2c\right)^2}=\frac{4}{a+b+2c}\end{cases}}\)

\(\Rightarrow VT\ge\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\)

Xét \(\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\)

Áp dụng bất đẳng thức cộng mẫu số 

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

\(=\frac{36}{4\left(a+b+c\right)}=\frac{36}{12}=3\)

Mà \(VT\ge\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\)

\(\Rightarrow VT\ge3\)

\(\Leftrightarrow\frac{a+3}{3a+bc}+\frac{b+3}{3b+ca}+\frac{c+3}{3c+ab}\ge3\left(đpcm\right)\)

Chúc bạn học tốt !!!

+ thêm bớt bc,ca,ab lần lượt cho P ta được

\(P=\frac{a^3}{3a+3bc-\left(ab+ac+bc\right)}+\frac{b^3}{3b+3ca-\left(ab+ac+bc\right)}+\frac{c^3}{3c+3ab-\left(ab+ac+bc\right)}+3abc\)

áp dụng BDT cô si cho mẫu ta có

\(3a+3bc\ge2\sqrt{9abc}=6\sqrt{abc}\)

suy ra

\(\frac{a^3}{3a+3bc-\left(ab+ac+bc\right)}\le\frac{a^3}{6\sqrt{abc}-\left(ab+ac+Bc\right)}\)

tương tự với các BDT còn lại suy ra :

\(P\le\frac{a^3}{6\sqrt{abc}-\left(ab+ac+bc\right)}+\frac{b^3}{6\sqrt{abc}-\left(ab+ac+bc\right)}+\frac{c^3}{6\sqrt{abc}-\left(ab+ac+bc\right)}+3abc\)

đên đây easy chưa ? chung mẫu + lại với nhau ta được

\(P\le\frac{a^3+b^3+c^3}{6\sqrt{abc}-\left(ab+ac+bc\right)}+3abc\)

áp dụng BDT cô si ta có

\(ab+bc+ca\le a^2+b^2+c^2\) luôn đúng thay vào ta được

ta có   \(a^2+B^2+c^2=\left(a+b+c\right)^2-2\left(ab+bc+ca\right)\) thêm bớt + hằng đẳng thức

thay vào và đổi dấu ta được

\(P\le\frac{a^3+b^3+c^3}{6\sqrt{abc}-9+2\left(ab+bc+Ca\right)}+3abc\)

có  \(ab+1\ge2\sqrt{ab}\)

\(ca+1\ge2\sqrt{ac}\)

\(bc+1\ge2\sqrt{bc}\)

\(\Rightarrow2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\le ab+bc+ca+3\)

ta lại có

\(\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\le a+B+c\left(cosi\right)\) suy ra

\(2\left(a+b+c\right)\le ab+bc+ca+3\Leftrightarrow6\le ab+Bc+ca+3\Leftrightarrow ab+bc+ca\ge3\)

  suy ra  

\(P\le\frac{\left(a^3+b^3+c^3\right)}{6\sqrt{abc}-9+2\left(3\right)}=\frac{\left(a^3+b^3+c^3\right)}{6\sqrt{abc}-3}\)

\(P\le\frac{\left(a^3+b^3+c^3\right)}{6\sqrt{abc}-3}+3abc\)

ta có

\(a.a.a\le\frac{\left(a+a+a\right)^3}{27}\)

\(b.b.b\le\frac{\left(b+b+b\right)^3}{27}\)

\(c.c.c\le\frac{\left(c+c+C\right)^3}{27}\)

\(a^3+b^3+c^3\le\frac{\left(3a\right)^3+\left(3b\right)^3+\left(3c\right)^3}{27}\)

bạn ơi chắc là đề sai rồi làm sao có thể đi chứng minh được cái

\(a^3+b^3+c^3\le a+b+c\) 

bạn xem lại đi nha @@

Ta có: \(\frac{5a^3-b^3}{ab+3a^2}=\frac{3a^3-b^3}{ab+3a^2}+\frac{2a^3}{ab+3a^2}\)

\(=a-\frac{a^2b+b^3}{ab+3a^2}+\frac{2a^3}{ab+3a^2}\)

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

Áp dụng BĐT AM - GM ( x² + y² \(\ge2xy\)) ta có:

(1) \(\le a-\frac{2ab^2}{a\left(b+3a\right)}+\frac{2a^2}{b+3a}\) = \(a-\frac{2b^2}{b+3a}+\frac{2a^2}{b+3a}\) (2)

Tương tự ta cũng có:

\(\frac{5b^3-c^3}{bc+3b^2}\le b-\frac{2c^2}{c+3b}+\frac{2b^2}{c+3b}\left(3\right)\)

\(\frac{5c^3-a^2}{ca+3c^2}\)\(\le c-\frac{2a^2}{a+3c}+\frac{2c^2}{a+3c}\)(4)

Từ (2), (3), (4) \(\Rightarrow\frac{5a^3-b^3}{ab+3a^2}+\frac{5b^3-c^3}{bc+3b^2}+\frac{5c^3-a^3}{ca+3c^2}\le a+b+c+\left(\frac{2a^2}{a+3c}-\frac{2a^2}{a+3c}\right)+\left(\frac{2b^2}{b+3c}-\frac{2b^2}{b+3c}\right)+\left(\frac{2c^2}{c+3a}-\frac{2c^2}{c+3a}\right)=a+b+c\le2018\)

Vậy \(\frac{5a^3-b^3}{ab+3a^2}+\frac{5b^3-c^3}{bc+3b^2}+\frac{5c^3-a^3}{ca+3c^2}\le2018\)

1.

\(P=\frac{a^4}{abc}+\frac{b^4}{abc}+\frac{c^4}{abc}\ge\frac{\left(a^2+b^2+c^2\right)^2}{3abc}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)\left(a+b+c\right)}{3abc\left(a+b+c\right)}\)

\(P\ge\frac{\left(a^2+b^2+c^2\right).3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}}{3abc\left(a+b+c\right)}=\frac{3\left(a^2+b^2+c^2\right)}{a+b+c}\)

Dấu "=" khi \(a=b=c\)

2.

\(P=\sum\frac{a^2}{ab+2ac+3ad}\ge\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\ge\frac{\left(a+b+c+d\right)^2}{4.\frac{3}{8}\left(a+b+c+d\right)^2}=\frac{2}{3}\)

Dấu "=" khi \(a=b=c=d\)

Thục Trinh, tran nguyen bao quan, Phùng Tuệ Minh, Ribi Nkok Ngok, Lê Nguyễn Ngọc Nhi, Tạ Thị Diễm Quỳnh,

Nguyễn Huy Thắng, ?Amanda?, saint suppapong udomkaewkanjana

Help me!

Bài thứ hai đó áp dụng bđt cauchy showas là ra rồi sử dụng tch bắc cầu tệ.

Xét Bất đẳng thức phụ:

\(\frac{5b^3-a^3}{ab+3b^2}\le2b-a\Leftrightarrow5b^3-a^3\le\left(2b-a\right)\left(ab+3b^2\right)\)

\(\Leftrightarrow a^2b+ab^2\le a^3+b^3\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\) (luôn đúng)

Tương tự ta có:

\(\frac{5a^3-b^3}{ab+3a^2}\le2a-c\);\(\frac{5c^3-a^3}{ac+3c^2}\le2c-b\)

Cộng lại theo vế ta có:

\(\frac{5a^3-b^3}{ab+3a^2}+\frac{5b^3-c^3}{bc+3b^2}+\frac{5c^3-a^3}{ac+3c^2}\le2b-a+2a-c+2c-b=a+b+c=2007\)

Đpcm

pịa pịa pịa 

l405ttol9to5l9g

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 )

 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 \(\Leftrightarrow a=b=c=1\)

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 )