\(ab+bc+ca=3abc\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\) (do a,b,c là các số dương)

Áp dụng BĐT Bunhiacopxki dạng phân thức:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{c}\ge\dfrac{6^2}{a+2b+3c}\)

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

Tương tự: \(\left\{{}\begin{matrix}\dfrac{36}{b+2c+3a}\le\dfrac{1}{b}+\dfrac{2}{c}+\dfrac{3}{a}\left(2\right)\\\dfrac{36}{c+2a+3b}\le\dfrac{1}{c}+\dfrac{2}{a}+\dfrac{3}{b}\left(3\right)\end{matrix}\right.\)

Lấy (1) + (2) + (3) ta được:

\(36F\le6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=6.3=18\)

\(\Rightarrow F\le\dfrac{1}{2}\)

MaxF=1/2 khi \(a=b=c=1\)

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 \(a=b=1\)

Ta có:\(\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ac}\ge\dfrac{9}{1+1+1+ab+bc+ca}\)(AM-GM)

Lại có:\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)

\(\Rightarrow\dfrac{9}{3+ab+bc+ca}\ge\dfrac{9}{3+a^2+b^2+c^2}=\dfrac{9}{6}=\dfrac{3}{2}\)

\(\Rightarrowđpcm\)

Cháu làm cho bác câu 2 thôi,câu 3 THANGDZ làm rồi sợ mất bản quyền lắm:v

Lời giải:

Áp dụng liên tiếp bất đẳng thức AM-GM và Cauchy-Schwarz ta có:

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

\(=\dfrac{a^2}{a^2+2ab+3ac}+\dfrac{b^2}{b^2+2bc+3ab}+\dfrac{c^2}{c^2+2ac+3bc}\)

\(\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+5ab+5bc+5ac}\)

\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+3\left(ab+bc+ac\right)}\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\left(a+b+c\right)^2}=\dfrac{1}{2}\)

\(ab+bc+ca=abc\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

Đặt vế trái của BĐT cần chứng minh là P

Ta có:

\(\dfrac{1}{a+2b+3c}=\dfrac{1}{a+b+b+c+c+c}\le\dfrac{1}{6^2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{c}\right)\)

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

Tương tự:

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

Cộng vế:

\(P\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{6}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)

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

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

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

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

ms lớp 9 thôi nha bạn

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

\(=\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)

Tương tự:

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

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

Cộng vế:

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

\(P\le\dfrac{1}{9}.\left(a+b+c+\dfrac{a+b+c}{2}\right)=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Bài 1:

Biểu thức chỉ có giá trị lớn nhất, không có giá trị nhỏ nhất.

\(P=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}=1-\frac{1}{x+1}+1-\frac{1}{y+1}+1-\frac{1}{z+1}\)

\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)

Giờ chỉ cần cho biến $x$ nhỏ vô cùng đến $0$, khi đó giá trị biểu thức trong ngoặc sẽ tiến đến dương vô cùng, khi đó P sẽ tiến đến nhỏ vô cùng, do đó không có min

Nếu chuyển tìm max thì em tìm như sau:

Áp dụng BĐT Cauchy_Schwarz:

\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\geq \frac{(1+1+1)^2}{x+1+y+1+z+1}=\frac{9}{x+y+z+3}=\frac{9}{4}\)

Do đó: \(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\leq 3-\frac{9}{4}=\frac{3}{4}\)

Vậy \(P_{\min}=\frac{3}{4}\Leftrightarrow x=y=z=\frac{1}{3}\)

Bài 2:

Áp dụng BĐT Cauchy-Schwarz :

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

\(\Rightarrow \frac{ab}{a+3b+2c}\leq \frac{1}{9}\left(\frac{ab}{a+c}+\frac{ab}{b+c}+\frac{a}{2}\right)\)

Hoàn toàn tương tự:

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

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

Cộng theo vế:

\(\Rightarrow \text{VT}\leq \frac{1}{9}\left(\frac{b(a+c)}{a+c}+\frac{a(b+c)}{b+c}+\frac{c(a+b)}{a+b}+\frac{a+b+c}{2}\right)\)

hay \(\text{VT}\leq \frac{a+b+c}{6}\) (đpcm)

Dấu bằng xảy ra khi $a=b=c$

Bài 3:

Áp dụng BĐT AM-GM:

\(P=\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\)

\(P=\frac{1}{(a^2+b^2)+(b^2+1)+2}+\frac{1}{(b^2+c^2)+(c^2+1)+2}+\frac{1}{(c^2+a^2)+(a^2+1)+2}\)

\(\leq \frac{1}{2ab+2b+2}+\frac{1}{2bc+2c+1}+\frac{1}{ac+2a+2}\)

\(\Leftrightarrow P\leq \frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ac+a+1}\right)\)

Gọi biểu thức trong ngoặc lớn là $N$

Do $abc=1$ nên ta có:

\(N=\frac{ac}{ab.ac+b.ac+ac}+\frac{a}{bc.a+c.a+a}+\frac{1}{ac+a+1}\)

\(=\frac{ac}{a+1+ac}+\frac{a}{1+ca+a}+\frac{1}{ca+a+1}=\frac{ac+a+1}{ac+a+1}=1\)

DO đó: \(P\leq \frac{1}{2}N=\frac{1}{2}\)

Vậy \(P_{\max}=\frac{1}{2}\). Dấu bằng xảy ra khi $a=b=c=1$

Lời giải:

Áp dụng BĐT Cauchy-Schwarz:

\(T=\frac{\frac{1}{a^2}}{\frac{1}{b}+\frac{1}{c}}+\frac{\frac{1}{b^2}}{\frac{1}{c}+\frac{1}{a}}+\frac{\frac{1}{c^2}}{\frac{1}{a}+\frac{1}{b}}\geq \frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2}{2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}=\frac{1}{2}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\)

\(\geq \frac{1}{2}.3\sqrt[3]{\frac{1}{abc}}=\frac{3}{2}\) (theo BĐT AM-GM)

Vậy $T_{\min}=\frac{3}{2}$.

Giá trị này đạt tại $a=b=c=1$