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`sqrta+1>sqrt{a+1}`
`<=>a+2sqrta+1>a+1`
`<=>2sqrta>0`
`<=>sqrta>0AAa>0`
`sqrt{a-1}<sqrta`
`<=>a-1<a`
`<=>-1<0` luôn đúng
`sqrt6-1>sqrt3-sqrt2`
`<=>sqrt6-sqrt3+sqrt2-1>0`
`<=>sqrt3(sqrt2-1)+sqrt2-1>0`
`<=>(sqrt2-1)(sqrt3+1)>0` luôn đúng
![](https://rs.olm.vn/images/avt/0.png?1311)
\(https://scontent.fhph1-1.fna.fbcdn.net/v/t34.0-12/19987311_122536408488931_1351154453_n.jpg?oh=553755e5363013e1853ab6f5ed63a600&oe=59BF5CA7\)https://scontent.fhph1-1.fna.fbcdn.net/v/t34.0-12/19987311_122536408488931_1351154453_n.jpg?oh=553755e5363013e1853ab6f5ed63a600&oe=59BF5CA7
Ấn vào linh đấy ế
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\(M=\left(\dfrac{1}{2+2\sqrt{a}}+\dfrac{1}{2-2\sqrt{a}}-\dfrac{a^2+1}{1-a^2}\right)\)
\(=\left(\dfrac{1}{2\left(1+\sqrt{a}\right)}+\dfrac{1}{2\left(1-\sqrt{a}\right)}-\dfrac{a^2+1}{1-a^2}\right).\dfrac{1+a}{a}\)
\(=\left(\dfrac{1-\sqrt{a}+1+\sqrt{a}}{2\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\dfrac{a^2+1}{\left(1-a\right)\left(1+a\right)}\right).\dfrac{1+a}{a}\)
\(=\left(\dfrac{1}{1-a}-\dfrac{a^2+1}{\left(1-a\right)\left(1+a\right)}\right).\dfrac{1+a}{a}\)
\(=\dfrac{1+a-a^2-1}{\left(1-a\right)\left(1+a\right)}.\dfrac{1+a}{a}\) (nghĩa là 1+a - (a^2 + 1 ) phá ngoặc thì đổi dấu như kia nhé.
quên mk chưa lm xong đã gửi r
\(=\dfrac{-a^2+a}{\left(1-a\right)\left(1+a\right)}.\dfrac{1+a}{a}\)
\(\dfrac{a\left(1-a\right)}{\left(1-a\right)\left(1+a\right)}.\dfrac{1+a}{a}=1\)( chia hết cho nhau thì = 1 nhé
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Đặt \(\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)=\left(x,y,z\right)\)
\(x+y+z\ge\frac{x^2+2xy}{2x+y}+\frac{y^2+2yz}{2y+z}+\frac{z^2+2zx}{2z+x}\)
\(\Leftrightarrow x+y+z\ge\frac{3xy}{2x+y}+\frac{3yz}{2y+z}+\frac{3zx}{2z+x}\)
\(\frac{3xy}{2x+y}\le\frac{3}{9}xy\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{3}\left(x+2y\right)\)
\(\Rightarrow\Sigma_{cyc}\frac{3xy}{2x+y}\le\frac{1}{3}\left[\left(x+2y\right)+\left(y+2z\right)+\left(z+2x\right)\right]=x+y+z\)
Dấu "=" xảy ra khi x=y=z
![](https://rs.olm.vn/images/avt/0.png?1311)
a.
Bình phương 2 vế, BĐT cần chứng minh trở thành:
\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)}+\sqrt{\left(b^2+1\right)\left(c^2+1\right)}+\sqrt{\left(c^2+1\right)\left(a^2+1\right)}\ge6\)
Ta có:
\(\sqrt{\left(a^2+1\right)\left(1+b^2\right)}\ge\sqrt{\left(a+b\right)^2}=a+b\)
Tương tự cộng lại:
\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)}+\sqrt{\left(b^2+1\right)\left(c^2+1\right)}+\sqrt{\left(c^2+1\right)\left(a^2+1\right)}\ge2\left(a+b+c\right)=6\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
b.
\(\sum\dfrac{a+1}{a^2+2a+3}=\sum\dfrac{a+1}{a^2+1+2a+2}\le\sum\dfrac{a+1}{4a+2}\)
Nên ta chỉ cần chứng minh:
\(\sum\dfrac{a+1}{4a+2}\le1\Leftrightarrow\sum\dfrac{4a+4}{4a+2}\le4\)
\(\Leftrightarrow\sum\dfrac{1}{2a+1}\ge1\)
Đúng đo: \(\dfrac{1}{2a+1}+\dfrac{1}{2b+1}+\dfrac{1}{2c+1}\ge\dfrac{9}{2\left(a+b+c\right)+3}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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Em tham khảo tại link dưới đây:
Câu hỏi của Nguyen Dinh Dung - Toán lớp 8 - Học toán với OnlineMath