\(\dfrac{1}{\left(a+b+a+c\right)^2}\le\dfrac{1}{4\left(a+b\right)\left(a+c\right)}=\dfrac{1}{4\left(a^2+ab+bc+ca\right)}\le\dfrac{1}{64}\left(\dfrac{1}{a^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)

\(\le\dfrac{1}{64}\left(\dfrac{1}{a^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=\dfrac{1}{64}\left(\dfrac{2}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\)

Tương tự và cộng lại:

\(P\le\dfrac{1}{64}\left(\dfrac{4}{a^2}+\dfrac{4}{b^2}+\dfrac{4}{c^2}\right)=\dfrac{1}{16}.3=\dfrac{3}{16}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Áp dụng bđt: \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(1\right)\)

\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\)

\(\Rightarrow P\le\dfrac{1}{16}\left[\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)^2+\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)^2+\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)^2\right]\)\(\Rightarrow16P\le\dfrac{2}{\left(a+b\right)^2}+\dfrac{2}{\left(b+c\right)^2}+\dfrac{2}{\left(a+c\right)^2}+\dfrac{2}{\left(a+b\right)\left(b+c\right)}+\dfrac{2}{\left(a+b\right)\left(b+c\right)}+\dfrac{2}{\left(b+c\right)\left(c+a\right)}\)

Áp dụng: \(x^2+y^2+z^2\ge xy+yz+xz\left(2\right)\) với a+b=x,b+c=y,c+a=z

\(\Rightarrow16P\le\dfrac{4}{\left(a+b\right)^2}+\dfrac{4}{\left(b+c\right)^2}+\dfrac{4}{\left(c+a\right)^2}\)

Ta có: \(\dfrac{1}{\left(a+b\right)^2}\le4.16.\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2\)(do (1))

\(\Rightarrow16P\le\dfrac{1}{4}.16\left[\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2+\left(\dfrac{1}{b}+\dfrac{1}{c}\right)^2+\left(\dfrac{1}{c}+\dfrac{1}{a}\right)^2\right]=\dfrac{1}{4}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}\right)\le\dfrac{1}{4}.4.\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=3\)(do(2) và \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=3\))

\(\Rightarrow P\le\dfrac{3}{16}\)

\(ĐTXR\Leftrightarrow a=b=c=1\)

                      Bài làm :

Ta có :

\(\left(a+b\right)^2\ge4ab\)

\(\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)

\(\Leftrightarrow\frac{4}{a+b}\le\frac{1}{a}+\frac{1}{b}\)

\(\Leftrightarrow\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\left(1\right)\)

Dấu "=" xảy ra khi : a=b

Chứng minh tương tự như trên ; ta có :

\(\hept{\begin{cases}\frac{1}{b+c}\text{≤}\frac{1}{4}\left(\frac{1}{b}+\frac{1}{c}\right)\left(2\right)\\\frac{1}{c+a}\text{≤}\frac{1}{4}\left(\frac{1}{c}+\frac{1}{a}\right)\left(3\right)\end{cases}}\)

Cộng vế với vế của (1) ; (2) ; (3) ; ta được :

\(A\text{≤}\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\text{=}\frac{3}{2}\)

Dấu "=" xảy ra khi ;

\(\hept{\begin{cases}a=b=c\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\end{cases}}\Leftrightarrow a=b=c=1\)

Vậy Max (A) = 3/2 khi a=b=c=1

quản lí tên kiểu j z

Trước tiên ta đi chứng minh BĐT phụ là:

Với $a,b>0$ thì $a^2+b^4\ge ab(a^2+b^2)$

Cách CM:

BĐT trên tương đương với: $(a-b{)}^2(a^2+ab+b^2)\ge 0$ (luôn đúng)

Quay trở về bài toán chính: Áp dụng BĐT phụ trên :

$\Rightarrow \frac{c}{a^4+b^4+c}\le \frac{c}{ab(a^2+b^2)+c^{2}ab}=\frac{c}{ab(a^2+b^2+c^2)}=\frac{c^2}{a^2+b^2+c^2}$

Thực hiện tương tự với các phân thức còn lại và cộng theo vế:

$\Rightarrow T\le \frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1$ (đpcm)

Dấu bằng xảy ra khi a=b=c=1

Nó bị mất cái dấu gạch ngang chỗ phân số nha b

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

=> \(\dfrac{1}{a+1}=1-\dfrac{1}{b+1}+1-\dfrac{1}{c+1}=\dfrac{b}{b+1}+\dfrac{c}{c+1}\ge2\sqrt{\dfrac{bc}{\left(b+1\right)\left(c+1\right)}}\)( AM-GM)

Tương tự ta có \(\dfrac{1}{b+1}\ge2\sqrt{\dfrac{ac}{\left(a+1\right)\left(c+1\right)}}\); \(\dfrac{1}{c+1}\ge2\sqrt{\dfrac{ab}{\left(a+1\right)\left(b+1\right)}}\)

Nhân vế với vế các bđt trên

=> \(\dfrac{1}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge8\sqrt{\dfrac{a^2b^2c^2}{\left(a+1\right)^2\left(b+1\right)^2\left(c+1\right)^2}}=8\cdot\dfrac{abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\)

=> \(1\le8abc\)<=> \(abc\le\dfrac{1}{8}\)

Đẳng thức xảy ra <=> a=b=c=1/2

ý quên thiếu KL

Vậy MaxP = 1/8 <=> a=b=c=1/2

\(\dfrac{1}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{b+c}\ge\dfrac{16}{2a+3b+3c}\)

\(\dfrac{1}{b+c}+\dfrac{1}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{a+c}\ge\dfrac{16}{2b+3a+3c}\)

\(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{a+b}+\dfrac{1}{a+b}\ge\dfrac{16}{2c+3a+3b}\)

cộng tất cả lại ta được \(4.2017\ge16.\left(\dfrac{1}{2a+3b+3c}+\dfrac{1}{2b+3a+3c}+\dfrac{1}{2c+3a+3b}\right)< =>P\le\dfrac{2017}{4}\)

dấu bằng xảy ra khi \(\left\{{}\begin{matrix}\dfrac{1}{a+b}=\dfrac{1}{b+c}=\dfrac{1}{a+c}\\\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}=2017\end{matrix}\right.< =>\left\{{}\begin{matrix}a=b=c\\\dfrac{3}{2a}=\dfrac{3}{2b}=\dfrac{3}{2c}=2017\end{matrix}\right.< =>a=b=c=\dfrac{3}{4034}}\)

mấy cái bất đẳng thức ở đầu là như nào v ạ

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

\(\Rightarrow P\le\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Thỏa mãn $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ hay $a+b+c=1$ vậy bạn?