Chính bài của em:

Cho \(a,b,c\ge1\). CMR: \(a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)+2\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}... - Hoc24

do \(a,b,c\ge1\)\(=>\left\{{}\begin{matrix}b+c\ge2\\c+a\ge2\\a+b\ge2\end{matrix}\right.\)

\(=>\left\{{}\begin{matrix}a\left(b+c\right)\ge2a\\b\left(c+a\right)\ge2b\\c\left(a+b\right)\ge2c\end{matrix}\right.\)

\(=>\) biểu thức đề bài cho\(\ge2\left(a+b+c+\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\right)\)

\(2\left(1+1+1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}\right)=9\)

dấu= xảy ra<=>a=b=c=1

\(a;b;c\ge1\Rightarrow\left\{{}\begin{matrix}ab;bc;ca\ge1\\ab+bc+ca\ge3\end{matrix}\right.\)

Ta có:

\(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}=\dfrac{a^2+b^2+2}{a^2b^2+a^2+b^2+1}=1-\dfrac{a^2b^2-1}{a^2b^2+a^2+b^2+1}\ge1-\dfrac{a^2b^2-1}{a^2b^2+2ab+1}=\dfrac{2}{ab+1}\)

Tương tự: \(\dfrac{1}{a^2+1}+\dfrac{1}{c^2+1}\ge\dfrac{2}{ac+1}\) ; \(\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\ge\dfrac{2}{bc+1}\)

Cộng vế: \(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\ge\dfrac{1}{ab+1}+\dfrac{1}{bc+1}+\dfrac{1}{ca+1}\)

Do đó: \(VT\ge2\left(ab+bc+ca+\dfrac{1}{ab+1}+\dfrac{1}{bc+1}+\dfrac{1}{ca+1}\right)\)

\(VT\ge2\left(ab+bc+ca+\dfrac{9}{ab+bc+ca+3}\right)\)

Đặt \(ab+bc+ca=x\ge3\Rightarrow VT\ge2\left(x+\dfrac{9}{x+3}\right)\)

\(VT\ge2\left(\dfrac{x+3}{4}+\dfrac{9}{x+3}+\dfrac{3x}{4}-\dfrac{3}{4}\right)\ge2\left(2\sqrt{\dfrac{9\left(x+3\right)}{4\left(x+3\right)}}+\dfrac{3}{4}.3-\dfrac{3}{4}\right)=9\)

a)Theo bất đẳng thức cauchy:

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

\(\Rightarrow\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge\dfrac{4}{a+b}.\left(a+b\right)\)

\(\Rightarrow\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\)

Dấu "=" xảy ra khi: \(a=b\)

Ta có điều phải chứng minh

b)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\)

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(a+b+c\right)\ge\dfrac{9}{a+b+c}.\left(a+b+c\right)\)

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(a+b+c\right)\ge9\)

Dấu "=" xảy ra khi:

\(a=b=c\)

Ta có điều phải chứng minh

c) Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có :

\(\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}\ge\dfrac{\left(1+1+1\right)^2}{A+B+C}=\dfrac{9}{A+B+C}\)

Dấu "=" xảy ra khi và chỉ khi\(\dfrac{1}{A}=\dfrac{1}{B}=\dfrac{1}{C}\)

Giải:

Áp dụng BĐT Cô si cho 3 số dương ta được:

\(a+b+c\ge3\sqrt[3]{abc}\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\)

Nhân theo vế 2 BĐT trên ta có:

\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\) (Đpcm)

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

Nếu đề là \(x,y,z\ge0\) thì làm như sau:

Áp dụng bất đẳng thức Cauchy ta có:

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

Đẳng thức xảy ra khi a = b = c

\(\Rightarrowđpcm\)

Nếu đề là \(a,b,c\ge0\) thì làm như sau:

Áp dụng bất đẳng thức Cauchy ta có:

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

Đẳng thức xảy ra khi a = b = c

\(\Rightarrowđpcm\)

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

=> (a+b).\(\left(\dfrac{1}{b}+\dfrac{1}{b}\right)\ge\left(a+b\right).\dfrac{4}{a+b}=4\left(dpcm\right)\)

b)\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{9}{a+b+c}\)

=>\(\left(a+b+c\right).\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\left(a+b+c\right).\dfrac{9}{a+b+c}=9\left(dpcm\right)\)

\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

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

\(=\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+3\)

Áp dụng BĐT Cô - si cho 2 số không âm:

\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)

\(\dfrac{a}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}=2\)

\(\dfrac{b}{c}+\dfrac{c}{b}\ge2\sqrt{\dfrac{b}{c}.\dfrac{c}{b}}=2\)

Suy ra:

\(\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{a}{c}+\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{c}{b}+3\ge2+2+2+3=9\)

Dấu "=" xảy ra khi: a = b = c

Áp dụng BĐT Cauchy dạng Engel , ta có :

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) ≥ \(\dfrac{9}{a+b+c}\)

⇔ \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\text{≥}\left(a+b+c\right).\dfrac{9}{a+b+c}\)

⇔ \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\text{≥}9\)

\("="\text{⇔}a=b=c\)

Áp dụng bất đẳng thức Cô - si cho 3 số không âm ta có :

\(a+b+c\ge3\sqrt[3]{abc}=3\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}=3\)

Nhân vế theo vế ta có :

\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)

Dấu " = " xảy ra khi a = b = c .

Cách 2:

Ta có:

\(A=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

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

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

\(=3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\)

Áp dụng BĐT AM-GM, ta có: \(\left\{{}\begin{matrix}\dfrac{a}{b}+\dfrac{b}{a}\ge2\\\dfrac{b}{c}+\dfrac{c}{b}\ge2\\\dfrac{c}{a}+\dfrac{a}{c}\ge2\end{matrix}\right.\)

=> \(A\ge9\)

P/s: Nhìn hơi dài nhưng trình bày ra thì không quá dài đâu! Ở đây mình làm hơi cẩn thận ::)))

Áp dụng Bất đẳng thức Côsi:

\(\left(a+b+c\right)\ge3\sqrt[3]{abc}\)

\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{\dfrac{1}{a}.\dfrac{1}{b}.\dfrac{1}{c}}\)

\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\dfrac{1}{a}.\dfrac{1}{b}.\dfrac{1}{c}}\)

\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)

Vậy \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\left(đpcm\right)\)

P/s: Ủa, đề này lớp 8 à? Sao cô mình lại cho bọn mình làm cái này nhỉ? WTF?????

Nguyễn thị kiều linh làm ko hiểu