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

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

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

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 .

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

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

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

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

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

Áp dụng BĐT Cô-si, ta có:

\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}\cdot\dfrac{b}{a}}\ge2\)

C/m tương tự, ta có:

\(\dfrac{a}{c}+\dfrac{c}{a}\ge2\)

\(\dfrac{b}{c}+\dfrac{c}{b}\ge2\)

\(\Rightarrow3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\ge2+2+2+3\)

\(\Rightarrow3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\ge9\)

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

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

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

1.

BĐT cần chứng minh tương đương:

\(\left(ab-1\right)\left(bc-1\right)\left(ca-1\right)\ge\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)\)

Ta có:

\(\left(ab-1\right)^2=a^2b^2-2ab+1=a^2b^2-a^2-b^2+1+a^2+b^2-2ab\)

\(=\left(a^2-1\right)\left(b^2-1\right)+\left(a-b\right)^2\ge\left(a^2-1\right)\left(b^2-1\right)\)

Tương tự: \(\left(bc-1\right)^2\ge\left(b^2-1\right)\left(c^2-1\right)\)

\(\left(ca-1\right)^2\ge\left(c^2-1\right)\left(a^2-1\right)\)

Do \(a;b;c\ge1\)  nên 2 vế của các BĐT trên đều không âm, nhân vế với vế:

\(\left[\left(ab-1\right)\left(bc-1\right)\left(ca-1\right)\right]^2\ge\left[\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)\right]^2\)

\(\Rightarrow\left(ab-1\right)\left(bc-1\right)\left(ca-1\right)\ge\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

Câu 2 em kiểm tra lại đề có chính xác chưa

2.

Câu 2 đề thế này cũng làm được nhưng khá xấu, mình nghĩ là không thể chứng minh bằng Cauchy-Schwaz được, phải chứng minh bằng SOS

Không mất tính tổng quát, giả sử \(c=max\left\{a;b;c\right\}\)

\(\Rightarrow\left(c-a\right)\left(c-b\right)\ge0\) (1)

BĐT cần chứng minh tương đương:

\(\dfrac{1}{a}-\dfrac{a+b}{bc+a^2}+\dfrac{1}{b}-\dfrac{b+c}{ac+b^2}+\dfrac{1}{c}-\dfrac{c+a}{ab+c^2}\ge0\)

\(\Leftrightarrow\dfrac{b\left(c-a\right)}{a^3+abc}+\dfrac{c\left(a-b\right)}{b^3+abc}+\dfrac{a\left(b-c\right)}{c^3+abc}\ge0\)

\(\Leftrightarrow\dfrac{c\left(b-a\right)+a\left(c-b\right)}{a^3+abc}+\dfrac{c\left(a-b\right)}{b^3+abc}+\dfrac{a\left(b-c\right)}{c^3+abc}\ge0\)

\(\Leftrightarrow c\left(b-a\right)\left(\dfrac{1}{a^3+abc}-\dfrac{1}{b^3+abc}\right)+a\left(c-b\right)\left(\dfrac{1}{a^3+abc}-\dfrac{1}{c^3+abc}\right)\ge0\)

\(\Leftrightarrow\dfrac{c\left(b-a\right)\left(b^3-a^3\right)}{\left(a^3+abc\right)\left(b^3+abc\right)}+\dfrac{a\left(c-b\right)\left(c^3-a^3\right)}{\left(a^3+abc\right)\left(c^3+abc\right)}\ge0\)

\(\Leftrightarrow\dfrac{c\left(b-a\right)^2\left(a^2+ab+b^2\right)}{\left(a^3+abc\right)\left(b^3+abc\right)}+\dfrac{a\left(c-b\right)\left(c-a\right)\left(a^2+ac+c^2\right)}{\left(a^3+abc\right)\left(c^3+abc\right)}\ge0\)

Đúng theo (1)

Dấu "=" xảy ra khi \(a=b=c\)

Áp dụng BĐT cauchy ta có:

\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}\cdot3\sqrt[3]{\dfrac{1}{abc}}=9\sqrt[3]{abc\cdot\dfrac{1}{abc}}=9\)

Dấu \("="\Leftrightarrow a=b=c\)

Bài 1a):

Ta có:

\(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\left(a+b\right).\dfrac{a+b}{ab}=\dfrac{a^2+2ab+b^2}{ab}=\dfrac{a^2+b^2}{ab}+2\)

Lại có: (a - b)² = a² - 2ab + b² \(\ge\) 0

\(\Rightarrow\) a² + b² \(\ge\) 2ab

\(\Rightarrow\) \(\dfrac{a^2+b^2}{ab}\ge2\)

\(\Rightarrow\) \(\dfrac{a^2+b^2}{ab}+2\ge4\)

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

Bài 2a):

Ta có: \(\left(\sqrt{a}-\sqrt{b}\right)^2=a-2\sqrt{ab}+b\ge0\)

\(\Rightarrow a+b\ge2\sqrt{ab}\)

Vậy ta có đpcm