Dễ thôi

Ta có:

\(ab+bc+ca+abc=4\Rightarrow\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1\) ( cái này cơ bản )

Theo AM - GM:

\(\left(a+b\right)^2+20=\left[\left(a+b\right)^2+4\right]+16\ge4\left(a+b\right)+16=4\left[\left(a+2\right)+\left(b+2\right)\right]\)

Áp dụng Cauchy Schwarz:

\(P\le\Sigma\frac{4}{4\left[\left(a+2\right)+\left(b+2\right)\right]}=\Sigma\frac{1}{\left(a+2\right)+\left(b+2\right)}\le\frac{1}{4}\Sigma\left(\frac{1}{a+2}+\frac{1}{b+2}\right)=\frac{1}{2}\)

Đẳng thức xảy ra tại a=b=c=1

Bác Cool kid chỉ em biến đối đê :D

Bài này có thể biểu diễn dưới dạng tổng bình phương nhưng khá xấu. (Vào TKHĐ xem ảnh)

\(\Leftrightarrow P=\dfrac{\sqrt{c-2}}{c}+\dfrac{\sqrt{a-3}}{a}+\dfrac{\sqrt{b-4}}{b}\)

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

\(dấu"="xảy\) \(ra\Leftrightarrow\left\{{}\begin{matrix}3=a-3\\4=b-4\\2=c-2\\\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=6\\b=8\\c=4\end{matrix}\right.\)

1)Từ đề bài:

`=>a^2+4b+4+b^2+4c+4+c^2+4a+4=0`

`<=>(a+2)^2+(b+2)^2+(c+2)^2=0`

`<=>a=b=c-2`

`ab+bc+ca=abc`

`<=>1/a+1/b+1/c=1`

`<=>(1/a+1/b+1/c)^2=1`

`<=>1/a^2+1/b^2+1/c^2+2/(ab)+2/(bc)+2/(ca)=1`

`<=>1/a^2+1/b^2+1/c^2=1-(2/(ab)+2/(bc)+2/(ca))`

`a+b+c=0`

Chia 2 vế cho `abc`

`=>1/(ab)+1/(bc)+1/(ca)=0`

`=>2/(ab)+2/(bc)+2/(ca)=0`

`=>1/a^2+1/b^2+1/c^2=1-0=1`

Có \(ab+bc+ac=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

Áp dụng các bđt sau:Với x;y;z>0 có: \(\dfrac{1}{x+y+z}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) và \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\) 

Có \(\dfrac{1}{a+3b+2c}=\dfrac{1}{\left(a+b\right)+\left(b+c\right)+\left(b+c\right)}\le\dfrac{1}{9}\left(\dfrac{1}{a+b}+\dfrac{2}{b+c}\right)\)\(\le\dfrac{1}{9}.\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{2}{c}\right)=\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{3}{b}+\dfrac{2}{c}\right)\)

CMTT: \(\dfrac{1}{b+3c+2a}\le\dfrac{1}{36}\left(\dfrac{1}{b}+\dfrac{3}{c}+\dfrac{2}{a}\right)\)

\(\dfrac{1}{c+3a+2b}\le\dfrac{1}{36}\left(\dfrac{1}{c}+\dfrac{3}{a}+\dfrac{2}{b}\right)\)

Cộng vế với vế => \(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{36}.6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)

Dấu = xảy ra khi a=b=c=3

Có \(a+b=2\Leftrightarrow2\ge2\sqrt{ab}\Leftrightarrow ab\le1\)

\(E=\left(3a^2+2b\right)\left(3b^2+2a\right)+5a^2b+5ab^2+2ab\)

\(=9a^2b^2+6\left(a^3+b^3\right)+4ab+5ab\left(a+b\right)+20ab\)

\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+4ab+5ab\left(a+b\right)+20ab\)

\(=9a^2b^2+48-18ab.2+4ab+5.2.ab+20ab\)

\(=9a^2b^2-2ab+48\)

Đặt \(f\left(ab\right)=9a^2b^2-2ab+48;ab\le1\), đỉnh \(I\left(\dfrac{1}{9};\dfrac{431}{9}\right)\)

Hàm đồng biến trên khoảng \(\left[\dfrac{1}{9};1\right]\backslash\left\{\dfrac{1}{9}\right\}\)

 \(\Rightarrow f\left(ab\right)_{max}=55\Leftrightarrow ab=1\)

\(\Rightarrow E_{max}=55\Leftrightarrow a=b=1\)

Vậy...

2,

\(ab\le\dfrac{1}{4}\left(a+b\right)^2=1\Rightarrow0\le ab\le1\)

\(E=9a^2b^2+6\left(a^3+b^3\right)+5ab\left(a+b\right)+24ab\)

\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+5ab\left(a+b\right)+24ab\)

\(=9a^2b^2-2ab+48\)

Đặt \(ab=x\Rightarrow0\le x\le1\)

\(E=9x^2-2x+48=\left(x-1\right)\left(9x+7\right)+55\le55\)

\(E_{max}=55\) khi \(x=1\) hay \(a=b=1\)

 Bạn tham khảo bài này trên Quanda nha.  

Lời giải:

Đặt \((a+1,b+1,c+1)=(x,y,z)\Rightarrow (a,b,c)=(x-1,y-1,z-1)\)

Khi đó:
\(ab+bc+ac+abc=2\)

\(\Leftrightarrow (x-1)(y-1)+(y-1)(z-1)+(z-1)(x-1)+(x-1)(y-1)(z-1)=2\)

\(\Leftrightarrow xyz-(x+y+z)+2=2\Leftrightarrow xyz=x+y+z\)

Vậy bài toán trở thành: Cho $x,y,z>0$ thỏa mãn \(x+y+z=xyz\)

Tìm max \(P=\sum \frac{x}{x^2+1}\)

----------------------------------

Ta có: \(x+y+z=xyz\Rightarrow x(x+y+z)=x^2yz\)

\(\Rightarrow x(x+y+z)+yz=yz(x^2+1)\)

\(\Leftrightarrow (x+y)(x+z)=yz(x^2+1)\Rightarrow x^2+1=\frac{(x+y)(x+z)}{yz}\)

Do đó: \(\frac{x}{x^2+1}=\frac{x}{\frac{(x+y)(x+z)}{yz}}=\frac{xyz}{(x+y)(x+z)}\)

\(\Rightarrow P=\sum \frac{x}{x^2+1}=\sum \frac{xyz}{(x+y)(x+z)}=\frac{2xyz(x+y+z)}{(x+y)(y+z)(x+z)}\)

Theo BĐT AM-GM:

\((x+y)(y+z)(x+z)=(x+y+z)(xy+yz+xz)-xyz\)

\(\geq (x+y+z).(xy+yz+xz)-\frac{(x+y+z)(xy+yz+xz)}{9}=\frac{8}{9}(x+y+z)(xy+yz+xz)\)

\(\Rightarrow P\leq \frac{2xyz(x+y+z)}{\frac{8}{9}(x+y+z)(xy+yz+xz)}=\frac{9}{4}.\frac{xyz}{xy+yz+xz}(*)\)

Mà: \((xy+yz+xz)^2\geq 3xyz(x+y+z)=3(xyz)^2\)

\(\Rightarrow xy+yz+xz\geq \sqrt{3}xyz(**)\)

Từ \((*);(**)\Rightarrow P\leq \frac{9}{4}.\frac{1}{\sqrt{3}}=\frac{3\sqrt{3}}{4}\). Vậy \(P_{\max}=\frac{3\sqrt{3}}{4}\)

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

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

Dấu "=" xảy ra khi \(a=b=c=3\)

Áp dụng cosi:

`a^2+bc>=2a\sqrt{bc}`

Hoàn toàn tương tự:

`=>P<=1/2(1/sqrt{ab}+1/sqrt{bc}+1/sqrt{ca})`

Áp dụng cosi:

`1/a+1/b+1/c>=1/sqrt(ab)+1/sqrt(bc)+1/sqrt(ca)`

`=>P<=1/2(1/a+1/b+1/c)`

`=>P<=1/2((ab+bc+ca)/(abc))<=(a^2+b^2+c^2)/(2(abc))=1/2`

Dấu "=" `<=>a=b=c=3`

\(\dfrac{1}{\sqrt{a^2-ab+b^2}}< =\dfrac{1}{\sqrt{2ab-ab}}=\dfrac{1}{\sqrt{ab}}\)

\(\sqrt{\dfrac{1}{b^2-bc+c^2}}< =\dfrac{1}{\sqrt{bc}};\sqrt{\dfrac{1}{c^2-ac+c^2}}< =\dfrac{1}{\sqrt{ac}}\)

=>P<=1/a+1/b+1/c=3

Dấu = xảy ra khi a=b=c=1

Đề bài có nhầm lẫn gì ko nhỉ?

\(T=\dfrac{ab}{a^2+b^2+ab}+\dfrac{bc}{b^2+c^2+2bc}+\dfrac{ca}{c^2+a^2+ca}\le\dfrac{ab}{2ab+ab}+\dfrac{bc}{2bc+bc}+\dfrac{ca}{2ca+ca}=1\)