1. Thực hiện phép tính:
\(\dfrac{1}{a\left(a-b\right)\left(a-c\right)}+\dfrac{1}{b\left(b-a\right)\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)\left(c-b\right)}\)
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1)\(\dfrac{c-b}{\left(a-b\right)\left(c-b\right)\left(a-c\right)}+\dfrac{a-c}{\left(b-a\right)\left(b-c\right)\left(a-c\right)}+\dfrac{b-a}{\left(b-a\right)\left(c-b\right)\left(c-a\right)}=\dfrac{c-b+a-c+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
\(=\dfrac{b-c}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}-\dfrac{a-c}{\left(b-c\right)\left(a-b\right)\left(a-c\right)}\)
\(+\dfrac{a-b}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=0\)
Ta có :
\(\dfrac{1}{a\left(a-b\right)\left(a-c\right)}+\dfrac{1}{b\left(b-a\right)\left(b-c\right)}\)
\(=\dfrac{1}{a\left(a-b\right)\left(a-c\right)}-\dfrac{1}{b\left(a-b\right)\left(b-c\right)}\)
\(=\dfrac{b\left(b-c\right)-a\left(a-c\right)}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{b^2-bc-a^2+ac}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{\left(b^2-a^2\right)-\left(bc-ac\right)}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{\left(b-a\right)\left(b+a\right)-c\left(b-a\right)}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{\left(b-a\right)\left(b+a-c\right)}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{\left(a-b\right)\left(-a-b+c\right)}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{-a-b+c}{ab\left(a-c\right)\left(b-c\right)}\)
Như vậy :
\(\dfrac{1}{a\left(a-b\right)\left(a-c\right)}+\dfrac{1}{b\left(b-a\right)\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)\left(c-b\right)}\)
\(=\dfrac{-a-b+c}{ab\left(a-c\right)\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)\left(c-b\right)}\)
\(=\dfrac{-a-b+c}{ab\left(a-c\right)\left(b-c\right)}+\dfrac{1}{c\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{c\left(-a-b+c\right)+ab}{abc\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{-ac-bc+c^2+ab}{abc\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{\left(-ac+ab\right)-\left(bc-c^2\right)}{abc\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{a\left(b-c\right)-c\left(b-c\right)}{abc\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{\left(a-c\right)\left(b-c\right)}{abc\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{1}{abc}\)
Sửa đề: \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
Ta có: \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
\(=\dfrac{c-a}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\dfrac{a-b}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\dfrac{b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\dfrac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
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 \(a=b=c\)
\(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)=8\)
\(\Leftrightarrow\dfrac{a+b}{a}\times\dfrac{b+c}{b}\times\dfrac{a+c}{c}=8\)
\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=8abc\)
~*~*~*~*~
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{a+c}\)
\(=\dfrac{3}{4}+\dfrac{ab}{\left(a+b\right)\left(b+c\right)}+\dfrac{bc}{\left(b+c\right)\left(c+a\right)}+\dfrac{ac}{\left(c+a\right)\left(a+b\right)}\) (1)
\(\Leftrightarrow\dfrac{a}{a+b}-\dfrac{ab}{\left(a+b\right)\left(b+c\right)}+\dfrac{b}{b+c}-\dfrac{bc}{\left(b+c\right)\left(c+a\right)}+\dfrac{c}{c+a}-\dfrac{ac}{\left(c+a\right)\left(a+b\right)}\)
\(=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{a}{a+b}\left(1-\dfrac{b}{b+c}\right)+\dfrac{b}{b+c}\left(1-\dfrac{c}{c+a}\right)+\dfrac{c}{a+c}\left(1-\dfrac{a}{a+b}\right)\)
\(=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{a}{a+b}\times\dfrac{c}{b+c}+\dfrac{b}{b+c}\times\dfrac{a}{a+c}+\dfrac{c}{a+c}\times\dfrac{b}{a+b}\)
\(=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{ac\left(a+c\right)+ab\left(a+b\right)+bc\left(b+c\right)}{\left(a+c\right)\left(b+c\right)\left(a+b\right)}=\dfrac{3}{4}\)
\(\Leftrightarrow ac\left(a+c\right)+ab\left(a+b\right)+bc\left(b+c\right)=\dfrac{3}{4}\times8abc\)
\(\Leftrightarrow ac\left(a+c\right)+ab\left(a+b\right)+bc\left(b+c\right)+2abc=8abc\)
\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=8abc\) luôn đúng
=> (1) đúng
Bạn cũng có thể giải bằng cách đặt \(x=\dfrac{a}{a+b};y=\dfrac{b}{b+c};z=\dfrac{c}{a+c}\).
\(\dfrac{1}{a\left(a-b\right)\left(a-c\right)}+\dfrac{1}{b\left(b-a\right)\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)\left(c-b\right)}\)
\(=\dfrac{1}{a\left(a-b\right)\left(a-c\right)}-\dfrac{1}{b\left(a-b\right)\left(b-c\right)}+\dfrac{1}{c\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{bc\left(b-c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\dfrac{ac\left(a-c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\dfrac{ab}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{b^2c-bc^2-a^{ 2}c+ac^2+a^2b-ab^2}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{\left(b-c\right)\left(a^2-b^2\right)-\left(a-b\right)\left(b^2-c^2\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{\left(b-c\right)\left(a-b\right)\left(a+b\right)-\left(a-b\right)\left(b-c\right)\left(b+c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{\left(b-c\right)\left(a-b\right)\left(a+b-b-c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{1}{abc}\)