Cho \(\frac{a}{b}>0\), chứng minh rằng \(\frac{a}{b}+\frac{b}{a}\ge2\)
Cho a ,b ,c ,d > 0 Chứng minh rằng : \(\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge2\)
Áp dụng BĐT \(\frac{1}{ab}\ge\frac{4}{\left(a+b\right)^2}\) với a , b > 0 ta có :
\(\frac{a}{b+c}+\frac{c}{d+a}=\frac{a\left(d+a\right)+c\left(b+c\right)}{\left(b+c\right)\left(d+a\right)}=\frac{ad+a^2+bc+c^2}{\left(b+c\right)\left(d+a\right)}\ge\frac{4\left(ad+a^2+bc+c^2\right)}{\left(a+b+c+d\right)^2}\) ( 1 )
\(\frac{b}{c+d}+\frac{d}{a+b}=\frac{b\left(a+b\right)+d\left(c+d\right)}{\left(a+b\right)\left(c+d\right)}=\frac{ab+b^2+cd+d^2}{\left(a+b\right)\left(c+d\right)}\ge\frac{4\left(ab+b^2+cd+d^2\right)}{\left(a+b+c+d\right)^2}\) ( 2 )
Từ ( 1 ) và ( 2 ) cộng theo từng vế:
\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge\frac{4\left(ab+bc+cd+ad+a^2+b^2+c^2+d^2\right)}{\left(a+b+c+d\right)^2}\)
Cần chứng minh rằng \(\frac{\left(ab+bc+cd+ad+a^2+b^2+c^2+d^2\right)}{\left(a+b+c+d\right)^2}\ge\frac{1}{2}\)
\(\Rightarrow2\left(ab+bc+cd+ad+a^2+b^2+c^2+d^2\right)\ge\left(a+b+c+d\right)^2\)
\(\Rightarrow2ab+2bc+2cd+2ad+2a^2+2b^2+2c^2+2d^2\ge a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bc+2cd+2bd\)
\(\Rightarrow a^2+b^2+c^2+d^2\ge2ac+2bd\)
\(\Rightarrow a^2-2ac+c^2+b^2-2bd+d^2\ge0\)
\(\Rightarrow\left(a-c\right)^2+\left(b-d\right)^2\ge0\left(đpcm\right)\)
Vậy \(\frac{ab+bc+cd+ad+a^2+b^2+c^2+d^2}{\left(a+b+c+d\right)^2}\ge\frac{1}{2}\)
\(\Rightarrow\frac{4\left(ab+bc+cd+ad+a^2+b^2+c^2+d^2\right)}{\left(a+b+c+d\right)^2}\ge2\)
Vì \(\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge\frac{4\left(ab+bc+cd+ad+a^2+b^2+c^2+d^2\right)}{\left(a+b+c+d\right)^2}\)
Vậy \(\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge2\)
Cho a ,b ,c ,d > 0 Chứng minh rằng : \(\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge2\)
Áp dụng BĐT bunhiacopxki cho 2 bộ số \(\left(\sqrt{a}.\sqrt{b+c};\sqrt{b}.\sqrt{d+c};\sqrt{c}.\sqrt{d+a};\sqrt{d}.\sqrt{a+b}\right)\)
và \(\left(\frac{\sqrt{a}}{\sqrt{b+c}};\frac{\sqrt{b}}{\sqrt{d+c}};\frac{\sqrt{c}}{\sqrt{d+a}};\frac{\sqrt{d}}{\sqrt{a+b}}\right)\), ta được:
\(\left[a\left(b+c\right)+b\left(d+c\right)+c\left(d+a\right)+d\left(a+b\right)\right]\)\(\left(\frac{a}{b+c}+\frac{b}{d+c}+\frac{c}{a+d}+\frac{d}{a+b}\right)\)\(\ge\left(a+b+c+d\right)^2\)
\(\Leftrightarrow\frac{a}{b+c}+\frac{b}{d+c}+\frac{c}{a+d}+\frac{d}{a+b}\)\(\ge\frac{\left(a+b+c+d\right)^2}{ab+ac+bd+bc+cd+ac+ad+bd}\)(1)
Ta có \(\left(a+b+c+d\right)^2\ge2\left(ab+ac+bc+bd+cd+ac+ad+bd\right)\)
\(\Leftrightarrow\left(a-c\right)^2+\left(b-d\right)^2\ge0\)(luôn đúng)
Do đó: \(\left(a+b+c+d\right)^2\ge2\left(ab+ac+bc+bd+cd+ac+ad+bd\right)\)(2)
Từ (1) và (2) suy ra ĐPCM
Dấu "=" xảy ra khi và chỉ khi a=b=c=d
Áp dụng BĐT : \(\frac{1}{xy}\ge\frac{4}{\left(x+y\right)^2}\)với x,y > 0
Ta có : \(\frac{a}{b+c}+\frac{c}{d+a}=\frac{a^2+ad+bc+c^2}{\left(b+c\right)\left(a+d\right)}\ge\frac{4\left(a^2+ad+bc+c^2\right)}{\left(a+b+c+d\right)^2}\)
Tương tự : \(\frac{b}{c+d}+\frac{d}{a+b}\ge\frac{4\left(b^2+ab+cd+d^2\right)}{\left(a+b+c+d\right)^2}\)
\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge\frac{4\left(a^2+b^2+c^2+d^2+ad+bc+ab+cd\right)}{\left(a+b+c+d\right)^2}\)
Cần chứng minh : \(\frac{a^2+b^2+c^2+d^2+ad+bc+ab+cd}{\left(a+b+c+d\right)^2}\ge\frac{1}{2}\)
\(\Leftrightarrow2\left(a^2+b^2+c^2+d^2+ad+bc+ab+cd\right)\ge\left(a+b+c+d\right)^2\)
\(\Leftrightarrow\left(a-c\right)^2+\left(b-d\right)^2\ge0\)
Dấu "=" xảy ra khi a = c ; b = d
Vậy ....
Ta có: \(\frac{a}{x}+\frac{b}{y}\ge\frac{\left(a+b\right)^2}{xy}\)
Lại có: \(\frac{a}{b+c}+\frac{d}{a+b}\)
\(\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge\frac{\left(a+b+c+d\right)^2}{ab+bc+bc+bd+ca+cd+da+db}\)
Ta chứng minh: \(\left(a+b+c+d\right)^2\ge2\left(ab+ac+bc+bd+ca+cd+da+db\right)\)
\(\Leftrightarrow\left(a+c\right)^2+2\left(a+c\right)\left(b+d\right)+\left(b+d\right)^2\ge2\left(a+c\right)\left(b+d\right)+4ac+4bd\)
\(\Leftrightarrow\left(a+c\right)^2+\left(b+d\right)^2\ge4ac+4bd\)(đúng)
\(\Leftrightarrow\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge2\left(đpcm\right)\)
Dấu " = "xảy ra \(\Leftrightarrow a=b=c=d\)
Cho a,b,c > 0 . Chứng minh rằng: \(\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\ge2\left(\frac{a}{b+1}+\frac{b}{c+1}+\frac{c}{a+1}\right)\)
Theo BĐT AM-GM :
\(\sqrt{b}=\sqrt{b\cdot1}\le\frac{b+1}{2}\)
\(\Rightarrow\frac{a}{\sqrt{b}}\ge\frac{a}{\frac{b+1}{2}}=\frac{2a}{b+1}\)
Dấu "=" xảy ra \(\Leftrightarrow b=1\)
+ Tương tự ta cm đc :
\(\frac{b}{\sqrt{c}}\ge\frac{2b}{c+1}\). Dấu "=" xảy ra \(\Leftrightarrow c=1\)
\(\frac{c}{\sqrt{a}}\ge\frac{2c}{a+1}\). Dấu "=" xảy ra \(\Leftrightarrow a=1\)
Do đó : \(\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\ge2\left(\frac{a}{b+1}+\frac{b}{c+}+\frac{c}{a+1}\right)\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Cho phân số \(\frac{a}{b}\)với a,b >0. Chứng minh rằng:
\(\frac{a}{b}+\frac{b}{a}\ge2\)
Ta có : \(\frac{a}{b}+\frac{b}{a}-2\)
\(=\frac{a^2}{ab}+\frac{b^2}{ab}-\frac{2ab}{ab}\)
\(=\frac{a^2-2ab+b^2}{ab}\)
\(=\frac{\left(a-b\right)^2}{ab}\ge0\) ( do a;b > 0 )
Dấu "=" xảy ra khi :
\(a-b=0\Leftrightarrow a=b\)
Vậy ...
Áp dụng bđt AM-GM: \(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{ab}{ab}}=2\)
Dấu "=" xảy ra khi: a=b
Cho phân số \(\frac{a}{b}>0\), chứng minh rằng \(\frac{a}{b}+\frac{b}{a}\ge2\).
ưk,th1 và th2 đều cần thiết để chứng minh,đáng lẻ là có 1 trường hợp a<b nhưng mình cm là a>b rồi thì thôi
Bài 1: Cho a, b cùng dấu. Chứng minh rằng: \(\left(\frac{a^2+b^2}{2}\right)^3\le\left(\frac{a^3+b^3}{2}\right)^2\)
Bài 2: Cho \(a^2+b^2\ne0\). Chứng minh rằng: \(\frac{2ab}{a^2+4b^2}+\frac{b^2}{3a^2+2b^2}\le\frac{3}{5}\)
Bài 3: Cho a, b > 0. Chứng minh rằng: \(\frac{a}{b^2}+\frac{b}{a^2}+\frac{16}{a+b}\ge5\left(\frac{1}{a}+\frac{1}{b}\right)\)
Bài 4: Cho a, b>0. Chứng minh rằng: \(\frac{3a^2+2ab+3b^2}{a+b}\ge2\sqrt{2\left(a^2+b^2\right)}\)
cho a,b,c>0 và a+b+c=6
chứng minh rằng
\(\frac{a}{\sqrt{b^3+1}}+\frac{b}{\sqrt{c^3+1}}+\frac{c}{\sqrt{a^3+1}}\ge2\)
Áp dụng BĐT Cauchy cho các cặp số dương, ta có: \(VT=\Sigma\frac{a}{\sqrt{b^3+1}}=\Sigma\frac{a}{\sqrt{\left(b+1\right)\left(b^2-b+1\right)}}\)
\(\ge\Sigma\frac{a}{\frac{\left(b+1\right)+\left(b^2-b+1\right)}{2}}=\Sigma\frac{2a}{b^2+2}=\Sigma\left(a-\frac{ab^2}{b^2+2}\right)\)
\(=\Sigma\left(a-\frac{2ab^2}{b^2+b^2+4}\right)\ge\Sigma\left(a-\frac{2ab^2}{3\sqrt[3]{4b^4}}\right)\)\(=\Sigma\left[a-\frac{a\sqrt[3]{2b^2}}{3}\right]=\Sigma\left[a-\frac{a\sqrt[3]{2.b.b}}{3}\right]\)
\(\ge\Sigma\left[a-\frac{a\left(2+b+b\right)}{9}\right]\)\(=\left(a+b+c\right)-\frac{2\left(a+b+c\right)}{9}-\frac{2\left(ab+bc+ca\right)}{9}\)
\(=\frac{7\left(a+b+c\right)}{9}-\frac{2\left(ab+bc+ca\right)}{9}\)\(\ge\frac{7\left(a+b+c\right)}{9}-\frac{2.\frac{\left(a+b+c\right)^2}{3}}{9}=2\)
Đẳng thức xảy ra khi a = b = c = 2
chứng minh rằng nếu \(ab>0\) thì \(\frac{a}{b}+\frac{b}{a}\ge2\)
cho a,b,c,d>0 chứng minh \(\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{a+d}+\frac{d}{a+b}\ge2\)
lớp 6 làm thì hơi dài đấy, nếu bạn muốn thì có thể áp dụng các bất đẳng thức của lớp trên cho nhanh