Áp dụng BĐT bunhiacop ski dạng phân thức(cauchy schwart)

`=>A>=(a+b+c)^2/(a+b+b+c+a+c)`

`<=>A>=(a+b+c)^2/(2(a+b+c))=(a+b+c)/2`

Mà `a+b+c=6`

`=>A>=6/2=3`

Dấu "=" xảy ra khi `a=b=c=2`

Câu hỏi của Thu Nguyễn - Toán lớp 9 - Học trực tuyến OLM

tham khảo ^^

- Theo BĐT Cauchy ta có:

\(\sqrt{a.1}\le\dfrac{a+1}{2}\)

\(\sqrt{b.1}\le\dfrac{b+1}{2}\)

\(\sqrt{c.1}\le\dfrac{c+1}{2}\)

\(\sqrt{ab}\le\dfrac{a+b}{2}\)

\(\sqrt{bc}\le\dfrac{b+c}{2}\)

\(\sqrt{ca}\le\dfrac{c+a}{2}\)

\(\Rightarrow\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le\dfrac{3\left(a+b+c\right)+3}{2}=\dfrac{3.3+3}{2}=6\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Mà ta có: \(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=6\)

\(\Rightarrow a=b=c=1\)

\(M=\dfrac{a^{30}+b^4+c^{1975}}{a^{30}+b^4+c^{2023}}=\dfrac{1^{30}+1^4+1^{1975}}{1^{30}+1^4+1^{2023}}=1\)

ta co: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1.\)

=> a = b = c 

\(\Rightarrow S=\frac{4a-5b+2019c}{5a-5b+2020c}=\frac{4a-5a+2019a}{5a-5a+2020c}=\frac{2018a}{2020a}=\frac{1009}{1010}\)

ta co: a/b=b/c=c/a =  (a+b+c)/(b+c+a) = 1

=> a/b = 1 => a = b

b/c =  1 => b = c

=> a = b = c

\(\Rightarrow S=\frac{4a-5a+2019a}{5a-5a+2020a}=\frac{2018a}{2020a}=\frac{1009}{1010}.\)

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\)

\(\text{Suy ra :}\)

\(\frac{a}{b}=1\Leftrightarrow a=b\)(1)

\(\frac{b}{c}=1\Leftrightarrow b=c\)(2)

\(\text{Từ (1) và (2) suy ra }:\): \(a=b=c\)

\(S=\frac{4a-5b+2019c}{5a-5b+2020c}=\frac{4a-5a+2019a}{5a-5a+2020a}=\frac{2018a}{2020a}=\frac{1009}{1010}\)

>Chúc bạn học tốt<

Ta có: 

\(P^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\le\left(a+b+b+c+c+a\right)\left(1^2+1^2+1^2\right)\)

\(=6\left(a+b+c\right)=18\)

Suy ra \(P\le3\sqrt{2}\)

Dấu \(=\) xảy ra khi \(a=b=c=1\). 

grsgrsg

omgggggggggggggomgomgomgggomggomgo

????????????

Ta có: \(\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{b}\)

\(\Rightarrow bc+ca=2ca\)

\(P=\dfrac{a+b}{2a-b}+\dfrac{c+b}{2c-b}=\dfrac{ac+bc}{2ca-bc}+\dfrac{ca+ab}{2ca-ab}\)

\(=\dfrac{ca+bc}{ab}+\dfrac{ca+ab}{bc}=\dfrac{c}{b}+\dfrac{c}{a}+\dfrac{a}{b}+\dfrac{a}{c}=\dfrac{c+a}{b}+\dfrac{c}{a}+\dfrac{a}{c}\)

Ta có :

\(\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}\ge\dfrac{4}{a+c}\left(\text{Svácxơ}\right)\)\(\Rightarrow c+a\ge2b\)

Áp dụng bđt cô si cho 2 số dương

\(\dfrac{c}{a}+\dfrac{a}{c}\ge2\sqrt{\dfrac{c}{a}.\dfrac{a}{c}}=2\)

\(\Rightarrow P\ge\dfrac{2b}{b}+2=4\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)