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Ta có
\(A=\left(a-b+c\right)\left(\frac{1}{a}-\frac{1}{b}+\frac{1}{c}\right)=3+\left(\frac{a}{c}+\frac{c}{a}\right)-\left(\frac{a}{b}+\frac{b}{a}\right)-\left(\frac{b}{c}+\frac{c}{b}\right)\)
áp dụng bđt Cauchy ta có
\(A\ge3+2-2-2=1\)(đpcm)
Dấu "=" xảy ra khi a=b=c=1
\(\left(a-b+c\right)\left(\frac{1}{a}-\frac{1}{b}+\frac{1}{c}\right)\ge1\)
\(\Leftrightarrow\left(a-b\right)\left(b-c\right)\left(c+a\right)\ge0\)(đúng)
Vậy bài toán được chứng minh
Câu hỏi của Nguyễn Thiều Công Thành - Toán lớp 9 - Học toán với OnlineMath
Ta có :\(\left(a-\frac{1}{b}\right)\left(b-\frac{1}{c}\right)\left(c-\frac{1}{a}\right)\)
\(=\frac{ab-1}{b}.\frac{bc-1}{c}.\frac{ac-1}{a}\)
Ta lại có : \(\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)\left(c-\frac{1}{c}\right)\)
\(=\frac{a^2-1}{a}.\frac{b^2-1}{b}.\frac{c^2-1}{c}\)
Xí trước phần b
Ta có: \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\)
\(=\frac{abc}{a^3\left(b+c\right)}+\frac{abc}{b^3\left(c+a\right)}+\frac{abc}{c^3\left(a+b\right)}\)
\(=\frac{bc}{a^2b+ca^2}+\frac{ca}{b^2c+ab^2}+\frac{ab}{c^2a+bc^2}\)
\(=\frac{b^2c^2}{a^2b^2c+a^2bc^2}+\frac{c^2a^2}{ab^2c^2+a^2b^2c}+\frac{a^2b^2}{a^2bc^2+ab^2c^2}\)
\(=\frac{\left(bc\right)^2}{ab+ca}+\frac{\left(ca\right)^2}{bc+ab}+\frac{\left(ab\right)^2}{ca+bc}\)
\(\ge\frac{\left(bc+ca+ab\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\ge\frac{3\sqrt[3]{\left(abc\right)^2}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi: \(a=b=c=1\)
Cách làm khác của phần b ngắn gọn hơn:)
Ta có; \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\)
\(=\frac{\frac{1}{a^2}}{a\left(b+c\right)}+\frac{\frac{1}{b^2}}{b\left(c+a\right)}+\frac{\frac{1}{c^2}}{c\left(a+b\right)}\)
\(=\frac{\left(\frac{1}{a}\right)^2}{ab+ca}+\frac{\left(\frac{1}{b}\right)^2}{bc+ab}+\frac{\left(\frac{1}{c}\right)^2}{ca+bc}\)
\(\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(\frac{ab+bc+ca}{abc}\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\ge\frac{3\sqrt[3]{\left(abc\right)^2}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi: a = b = c = 1
1. Áp dụng BĐT Cauchy dạng Engle, ta có :
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\left(a+b+c\right)\left(\frac{9}{a+b+c}\right)\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
\(\frac{1}{3}\left(a^3+b^3+a+b\right)+ab\le a^2+b^2+1\)
\(\Leftrightarrow\frac{1}{3}\left(a+b\right)\left(a^2+b^2+1-ab\right)+ab\le a^2+b^2+1\)
\(\Leftrightarrow\left(a^2+b^2+1\right)\left(\frac{a+b}{3}-1\right)-ab\left(\frac{a+b}{3}-1\right)\le0\)
\(\Leftrightarrow\left(a^2+b^2+1-ab\right)\left(\frac{a+b}{3}-1\right)\le0\)
Vì a, b dương \(\Rightarrow a^2+b^2+1-ab>0\Rightarrow\left(\frac{a+b}{3}-1\right)\le0\Leftrightarrow a+b\le3\)
\(M=\frac{a^2+8}{a}+\frac{b^2+2}{b}=a+\frac{8}{a}+b+\frac{2}{b}=2a+2b+\frac{8}{a}+\frac{2}{b}-\left(a+b\right)\ge8+4-3=9\)
Áp dụng BĐT Cauchy cho a ; b dương
Dấu "=" xảy ra \(\Leftrightarrow a=2;b=1\)
b) Áp dụng BĐT Cauchy-schwarz ta có:
\(\frac{1}{1+3ab+a^2}+\frac{1}{1+3ab+b^2}\ge\frac{4}{2+a^2+2ab+b^2+4ab}\)\(=\frac{4}{2+\left(a+b\right)^2+4ab}\) (1)
Dấu " = " xảy ra <=> a=b=0,5
Áp dụng BĐT AM-GM ta có:
\(4ab=4.\sqrt{ab}.\sqrt{ab}\le\frac{4.\left(a+b\right)^2}{4}=\left(a+b\right)^2=1\)(2)
Dấu " = " xảy ra <=> a=b=0,5
Từ (1) và (2)
\(\Rightarrow\frac{1}{1+3ab+a^2}+\frac{1}{1+3ab+b^2}\ge\frac{4}{2+\left(a+b\right)^2+4ab\ge}\frac{4}{3+\left(a+b\right)^2}=\frac{4}{4}=1\)
Dấu " = " xảy ra <=> a=b=0,5
P/s : Làm siêu tắt
Ta có :
\(\left(1+\frac{a}{b}\right)^5+\left(1+\frac{b}{a}\right)^5\ge\left(1+\frac{a}{b}\right)\left(1+\frac{b}{a}\right)\left[\left(1+\frac{a}{b}\right)^3+\left(1+\frac{b}{a}\right)^3\right]\ge\left(1+\frac{a}{b}\right)^2\left(1+\frac{b}{a}\right)^2\left(2+\frac{a}{b}+\frac{b}{a}\right)=\frac{\left(a+b\right)^2.\left(a+b\right)^2}{a^2b^2}.\left(2+\frac{a}{b}+\frac{b}{a}\right)\ge\frac{4ab.4ab}{a^2b^2}.\left(2+2\right)=16.4=64\)
( AD BĐT phụ \(x^5+y^5\ge xy\left(x^3+y^3\right);x^3+y^3\ge xy\left(x+y\right)\) và BĐT Cô - si )
Dấu " = " xảy ra \(\Leftrightarrow a=b;a,b>0\)