cho a,b,c > 0 thỏa mãn a^2 +b^2 +c^2=1 .CMR: 1/1-ab +1/1-bc +1/1-ca =< 9/2
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\(\dfrac{\sqrt{ab+2c^2}}{\sqrt{1+ab-c^2}}=\dfrac{\sqrt{ab+2c^2}}{\sqrt{a^2+b^2+ab}}=\dfrac{ab+2c^2}{\sqrt{\left(a^2+b^2+ab\right)\left(ab+2c^2\right)}}\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\)
\(\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+a^2+b^2+2c^2}=\dfrac{ab+2c^2}{a^2+b^2+c^2}=ab+2c^2\)
Tương tự và cộng lại:
\(VT\ge ab+bc+ca+2\left(a^2+b^2+c^2\right)=2+ab+bc+ca\)
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Đặt A = \(\dfrac{a-b}{1+c^2}+\dfrac{b-c}{1+a^2}+\dfrac{c-a}{1+b^2}=0\)
= \(\dfrac{a-b}{c^2+ab+bc+ca}+\dfrac{b-c}{a^2+ab+bc+ca}+\dfrac{c-a}{b^2+ab+bc+ca}\)
= \(\dfrac{a-b}{\left(c+a\right)\left(c+b\right)}+\dfrac{b-c}{\left(a+b\right)\left(c+a\right)}+\dfrac{c-a}{\left(a+b\right)\left(b+c\right)}\)
= \(\dfrac{\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c+a\right)\left(c-a\right)}{\left(c+a\right)\left(b+c\right)\left(a+b\right)}\)
= \(\dfrac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
\(\dfrac{a-b}{1+c^2}+\dfrac{b-c}{1+a^2}+\dfrac{c-a}{1+b^2}\)
\(=\dfrac{a-b}{ab+bc+ca+c^2}+\dfrac{b-c}{ab+bc+ca+a^2}+\dfrac{c-a}{ab+bc+ca+b^2}\)
\(=\dfrac{a-b}{\left(c+a\right)\left(c+b\right)}+\dfrac{b-c}{\left(a+b\right)\left(a+c\right)}+\dfrac{c-a}{\left(b+a\right)\left(b+c\right)}\)
\(=\dfrac{\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\dfrac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
đặt \(A=\frac{1}{1-ab}+\frac{1}{1-bc}+\frac{1}{1-ca}\)
\(\Rightarrow A-3=P=\frac{ab}{1-ab}+\frac{bc}{1-bc}+\frac{ca}{1-ca}\)
áp dụng BĐT cô-si ta có:
\(a^2+b^2\ge2ab;b^2+c^2\ge2bc;c^2+a^2\ge2ca\)
\(\Rightarrow\frac{a^2+b^2}{2}\ge ab;\frac{b^2+c^2}{2}\ge bc;\frac{c^2+a^2}{2}\ge ca\)
\(\Rightarrow1-\frac{a^2+b^2}{2}\le1-ab;1-\frac{b^2+c^2}{2}\le1-bc;1-\frac{c^2+a^2}{2}\le1-ca\)
\(\Rightarrow P\le\frac{2ab}{\left(a^2+c^2\right)+\left(b^2+c^2\right)}+\frac{2bc}{\left(a^2+b^2\right)+\left(a^2+c^2\right)}+\frac{2ca}{\left(a^2+b^2\right)+\left(b^2+c^2\right)}\)
\(\Rightarrow P\le\frac{1}{2}\left(\frac{\left(a+b\right)^2}{\left(a^2+c^2\right)+\left(b^2+c^2\right)}+\frac{\left(b+c\right)^2}{\left(a^2+b^2\right)+\left(a^2+c^2\right)}+\frac{\left(c+a\right)^2}{\left(a^2+b^2\right)+\left(b^2+c^2\right)}\right)\)
Áp dụng BĐT Schwarts ta có:
\(\frac{\left(a+b\right)^2}{\left(a^2+c^2\right)+\left(b^2+c^2\right)}\le\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}\)
\(\frac{\left(b+c\right)^2}{\left(a^2+b^2\right)+\left(a^2+c^2\right)}\le\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}\)
\(\frac{\left(c+a\right)^2}{\left(a^2+b^2\right)+\left(b^2+c^2\right)}\le\frac{a^2}{a^2+b^2}+\frac{c^2}{b^2+c^2}\)
\(\Rightarrow P\le\frac{1}{2}\left(\frac{a^2+b^2}{a^2+b^2}+\frac{b^2+c^2}{b^2+c^2}+\frac{c^2+a^2}{c^2+a^2}\right)=\frac{1}{2}.3=\frac{3}{2}\)
\(\Rightarrow P+3\le\frac{3}{2}+3\)
\(\Rightarrow A\le\frac{9}{2}\)
dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
Bất đẳng thức cần chứng minh tương đương: \(\frac{1}{ab-1}+\frac{1}{bc-1}+\frac{1}{ca-1}\ge\frac{-9}{2}\)
Theo bất đẳng thức Bunyakovsky dạng phân thức, ta được: \(\frac{1}{ab-1}+\frac{1}{bc-1}+\frac{1}{ca-1}\ge\frac{9}{ab+bc+ca-3}\)
\(\ge\frac{9}{a^2+b^2+c^2-3}=\frac{9}{1-3}=\frac{-9}{2}\left(Q.E.D\right)\)
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
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Do \(abc=1\Rightarrow\) đặt \(\left(a;b;c\right)=\left(\dfrac{x}{y};\dfrac{y}{z};\dfrac{z}{x}\right)\)
\(VT=\dfrac{xz}{y\left(x+z\right)}+\dfrac{xy}{z\left(x+y\right)}+\dfrac{yz}{x\left(y+z\right)}=\dfrac{\left(xz\right)^2}{xyz\left(x+z\right)}+\dfrac{\left(xy\right)^2}{xyz\left(x+y\right)}+\dfrac{\left(yz\right)^2}{xyz\left(y+z\right)}\)
\(VT\ge\dfrac{\left(xy+yz+zx\right)^2}{2xyz\left(x+y+z\right)}\ge\dfrac{3xyz\left(x+y+z\right)}{2xyz\left(x+y+z\right)}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a^2+b^2+c^2=1\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=1+2\left(ab+bc+ca\right)\)
\(\Leftrightarrow\left(a+b+c\right)^2=1+2\left(ab+bc+ca\right)\)
\(\Rightarrow1+2\left(ab+bc+ca\right)\ge0\)
\(\Rightarrow ab+bc+ca\ge-\dfrac{1}{2}\)
Lại có:
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Leftrightarrow2ab+2bc+2ca\le2a^2+2b^2+2c^2\)
\(\Leftrightarrow ab+bc+ca\le a^2+b^2+c^2\)
\(\Leftrightarrow ab+bc+ca\le1\)
Cho a,b,c >0 thỏa mãn a+b+c= ab+bc+ca. CMR:
\(\frac{1}{a^2+b+1}+\frac{1}{b^2+c+1}+\frac{1}{c^2+a+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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Áp dụng BĐT Cauchy dạng phân thức :
\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\ge\frac{9}{ab+bc+ac}\)
\(\Rightarrow VT\ge\frac{1}{a^2+b^2+c^2}+\frac{9}{ab+bc+ac}\)
\(\Leftrightarrow VT\ge\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ac}+\frac{1}{ab+bc+ac}+\frac{7}{ab+ac+bc}\)
Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow ab+bc+ac\le\frac{1}{3}\left(a+b+c\right)^2=\frac{1}{3}\)
\(\Rightarrow\frac{7}{ab+bc+ac}\ge21\left(1\right)\)
Áp dụng bất đẳng thức Cauchy dạng phân thức
\(\Rightarrow\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ac}+\frac{1}{ab+bc+ac}\)
\(\ge\frac{9}{a^2+b^2+c^2+2\left(ab+bc+ac\right)}=9\) (2)
Từ (1) và (2)
\(\Rightarrow VT\ge21+9=30\left(đpcm\right)\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
Chúc bạn học tốt !!!
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Cho mk k nhé!
4/1x3x5 = 1/1x3 - 1/3x5
4/3x5x7 = 1/3x5 - 1/5x7
.............
A = 1/1x3 - 1/11x13
1/1x3x5 = 1/4 x (1/1x3 - 1/3x5)
1/3x5x7 = 1/4 x (1/3x5 - 1/5x7)
..........
B = 1/4 x (1/1x3 - 1/11x13)