Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT Cauchy cho 2 số dương ta được :
\(\dfrac{a^2}{b+3c}+\dfrac{b+3c}{16}\ge2\sqrt{\dfrac{a^2}{b+3c}\times\dfrac{b+3c}{16}}=\dfrac{2a}{4}\)
Suy ra \(\dfrac{a^2}{b+3c}\ge\dfrac{2a}{4}-\dfrac{b+3c}{16}\)
Cmtt ta cũng được :
\(\dfrac{b^2}{c+3a}\ge\dfrac{2b}{4}-\dfrac{c+3a}{16}\) \(\dfrac{c^2}{a+3b}\ge\dfrac{2c}{4}-\dfrac{a+3b}{16}\)
Khi đó :
\(\dfrac{a^2}{b+3c}+\dfrac{b^2}{c+3a}+\dfrac{c^2}{a+3b}\ge\dfrac{2a}{4}-\dfrac{b+3c}{16}+\dfrac{2b}{4}-\dfrac{c+3a}{16}+\dfrac{2c}{4}-\dfrac{a+3b}{16}\)
mà \(\dfrac{2a}{4}-\dfrac{b+3c}{16}+\dfrac{2b}{4}-\dfrac{c+3a}{16}+\dfrac{2c}{4}-\dfrac{a+3b}{16}=\dfrac{a+b+c}{4}\)
Vậy \(\dfrac{a^2}{b+3c}+\dfrac{b^2}{c+3a}+\dfrac{c^2}{a+3b}\ge\dfrac{a+b+c}{4}\)
Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Rightarrow\dfrac{a^2}{b+3c}+\dfrac{b^2}{c+3a}+\dfrac{c^2}{a+3b}\ge\dfrac{\left(a+b+c\right)^2}{4\left(a+b+c\right)}=\dfrac{a+b+c}{4}\) (đpcm)
Dấu " = " xảy ra khi \(a=b=c\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT Bunhiacopxki ta có:
\(\left(1+1+1\right)\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\)
Dấu " = " xảy ra <=> a=b=c=1
Có: \(ab+bc+ca\le\frac{1}{3}\left(a+b+c\right)^2\Leftrightarrow a+b+c\ge3\)( bạn tự c/m nhé )
Dấu " = " xảy ra <=> a=b=c
Áp dụng BĐT Cauchy-schwarz ta có:
\(\frac{a^4}{b+3c}+\frac{b^4}{c+3a}+\frac{c^4}{a+3b}\ge\frac{\left(a^2+b^2+c^2\right)^2}{4\left(a+b+c\right)}\ge\frac{\left[\frac{\left(a+b+c\right)^2}{3}\right]^2}{4\left(a+b+c\right)}=\frac{\left(a+b+c\right)^3}{36}\ge\frac{27}{36}=\frac{3}{4}\)
Dấu " = " xảy ra <=> a=b=c=1 ( bạn tự giải rõ ra nhé )
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT Bunhiacopxki ta có :
\(\left(1+1+1\right)\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)
Ta có : \(ab+bc+ca\le\frac{1}{3}\left(a+b+c\right)^2\Leftrightarrow a+b+c\ge3\) ( tự chứng minh ạ )
Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)
Áp dụng BĐT Cachy Schwarz ta có :
\(\frac{a^4}{b+3c}+\frac{b^4}{c+3a}+\frac{c^4}{a+3b}\ge\frac{\left(a^2+b^2+c^2\right)^2}{4\left(a+b+c\right)}\) \(\ge\frac{\left[\frac{\left(a+b+c\right)}{3}\right]^2}{4\left(a+b+c\right)}=\frac{\left(a+b+c\right)^3}{36}\)
\(\ge\frac{27}{36}=\frac{3}{4}\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\) ( bạn tự giải rõ ạ )
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a^3+a^3+b^3\ge3\sqrt[3]{a^6b^3}=3a^2b\)
\(b^3+b^3+c^3\ge3\sqrt[3]{b^6c^3}=3b^2c\)
\(c^3+c^3+a^3\ge3\sqrt[3]{c^6a^3}=3c^2a\)
Cộng vế theo vế có ngay điều phải chứng minh
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(a^4-5a^2+4=\)\(\left(a^4-4a^2\right)-\left(a^2-4\right)=a^2\left(a^2-4\right)-\left(a^2-4\right)=\left(a^2-1\right)\left(a^2-4\right)\)
\(=\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)\)
\(a^4-a^2+4a-4=a^2\left(a^2-1\right)+4\left(a-1\right)=a^2\left(a-1\right)\left(a+1\right)+4\left(a-1\right)\)
\(=\left(a-1\right)\left[a^2\left(a+1\right)+4\right]=\left(a-1\right)\left(a^3+a^2+4\right)\)
\(a^3+a^2+4=\left(a^3+2a^2\right)-\left(a^2+2a\right)+\left(2a+4\right)=a^2\left(a+2\right)-a\left(a+2\right)+2\left(a+2\right)\)
\(=\left(a^2-a+2\right)\left(a+2\right)\)
\(N=\frac{\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)}{\left(a-1\right)\left(a+2\right)\left(a^2-a+2\right)}=\frac{\left(a+1\right)\left(a-2\right)}{a^2-a+2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có : BĐT \(a^2+b^2\ge2ab\) = BĐT cauchuy .
-> Áp dụng BĐT cauchuy ta được :
\(\left\{{}\begin{matrix}a^4+b^4\ge2\sqrt{a^4b^4}=2a^2b^2\\c^4+d^4\ge2\sqrt{c^4d^4}=2c^2d^2\end{matrix}\right.\)
- Cộng 2 bpt lại ta được :
\(a^4+b^4+c^4+d^4\ge2a^2b^2+2c^2d^2=2\left(\left(ab\right)^2+\left(cd\right)^2\right)\)
- Mà \(\left(ab\right)^2+\left(cd\right)^2\ge2abcd\)
=> \(a^4+b^4+c^4+d^4\ge2.2abcd=4abcd\)
b, CMTT câu 1 .
- Áp dụng BĐT cauchuy ta được :
\(\left\{{}\begin{matrix}a^2+1\ge2a\\b^2+1\ge2b\\c^2+1\ge2c\end{matrix}\right.\)
- Nhân 3 bpt trên lại ta được :
\(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge2.2.2abc=8abc\)
hello