K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

2 tháng 4 2022

C/m : \(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}=1\) (*)

Thật vậy , (*) \(\Leftrightarrow\left(a+2\right)\left(b+2\right)+\left(b+2\right)\left(c+2\right)+\left(a+2\right)\left(c+2\right)=\left(a+2\right)\left(b+2\right)\left(c+2\right)\)

\(\Leftrightarrow ab+bc+ac+4\left(a+b+c\right)+12=abc+2\left(ab+bc+ac\right)+4\left(a+b+c\right)+8\)

\(\Leftrightarrow ab+bc+ac+abc=4\) (Đ)

=> (*) đúng ( đpcm ) 

2 tháng 4 2022

 Dạ , em đã hiểu rồi ạ!
Em cám ơn nhiều lắm ạ

NV
17 tháng 2 2022

Bài toán cơ bản:

\(abc=1\Rightarrow\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}=1\) 

Bunhiacopxki:

\(\left(a+b+c\right)\left(\dfrac{a}{\left(ab+a+1\right)^2}+\dfrac{b}{\left(bc+b+1\right)^2}+\dfrac{c}{\left(ac+c+1\right)^2}\right)\ge\left(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\right)^2=1\)

\(\Rightarrow\dfrac{a}{\left(ab+a+1\right)^2}+\dfrac{b}{\left(bc+b+1\right)^2}+\dfrac{c}{\left(ac+c+1\right)^2}\ge\dfrac{1}{a+b+c}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

17 tháng 2 2022

nhầm

NV
6 tháng 4 2022

\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{ab}{2b}\right)\)

\(=\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)

Tương tự:

\(\dfrac{bc}{b+3c+2a}\le\dfrac{1}{9}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}+\dfrac{b}{2}\right)\)

\(\dfrac{ac}{c+3a+2b}\le\dfrac{1}{9}\left(\dfrac{ac}{b+c}+\dfrac{ac}{a+b}+\dfrac{c}{2}\right)\)

Cộng vế:

\(P\le\dfrac{1}{9}\left(\dfrac{bc+ac}{a+b}+\dfrac{bc+ab}{a+c}+\dfrac{ab+ac}{b+c}+\dfrac{a+b+c}{2}\right)\)

\(P\le\dfrac{1}{9}.\left(a+b+c+\dfrac{a+b+c}{2}\right)=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

NV
6 tháng 3 2022

\(ab+bc+ca=abc\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

Đặt vế trái của BĐT cần chứng minh là P

Ta có:

\(\dfrac{1}{a+2b+3c}=\dfrac{1}{a+b+b+c+c+c}\le\dfrac{1}{6^2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{c}\right)\)

\(\Rightarrow\dfrac{1}{a+2b+3c}\le\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{3}{c}\right)\)

Tương tự:

\(\dfrac{1}{b+2c+3a}\le\dfrac{1}{36}\left(\dfrac{1}{b}+\dfrac{2}{c}+\dfrac{3}{a}\right)\) ; \(\dfrac{1}{c+2a+3b}\le\dfrac{1}{36}\left(\dfrac{1}{c}+\dfrac{2}{a}+\dfrac{3}{b}\right)\)

Cộng vế:

\(P\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{6}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)

NV
6 tháng 3 2022

Ta có:

\(x^4+y^4\ge\dfrac{1}{2}\left(x^2+y^2\right)^2=\dfrac{1}{2}\left(x^2+y^2\right)\left(x^2+y^2\right)\ge\dfrac{1}{2}.2xy\left(x^2+y^2\right)=xy\left(x^2+y^2\right)\)

Áp dụng:

\(P\le\dfrac{a}{a+bc\left(b^2+c^2\right)}+\dfrac{b}{b+ca\left(c^2+a^2\right)}+\dfrac{c}{c+ab\left(a^2+b^2\right)}\)

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

Dấu "=" xảy ra khi \(a=b=c=1\)

7 tháng 3 2022

Em cám ơn thầy đã dành thời gian giúp đỡ ạ!

 

6 tháng 3 2022

ca này để thầy lâm ròi:<

6 tháng 3 2022

:v

NV
6 tháng 3 2022

\(\dfrac{a}{a+2\sqrt{\left(a+bc\right)}}=\dfrac{a}{a+2\sqrt{a\left(a+b+c\right)+bc}}=\dfrac{a}{a+2\sqrt{\left(a+b\right)\left(a+c\right)}}\)

\(=\dfrac{a}{a+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}\)

\(\le\dfrac{a}{5^2}\left(\dfrac{1}{a}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}\right)\)

\(=\dfrac{a}{25}\left(\dfrac{1}{a}+\dfrac{8}{\sqrt{\left(a+b\right)\left(a+c\right)}}\right)=\dfrac{1}{25}+\dfrac{8}{25}.\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)

\(\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)

Tương tự:

\(\dfrac{b}{b+2\sqrt{b+ac}}\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)\)

\(\dfrac{c}{c+2\sqrt{c+ab}}\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)\)

Cộng vế:

\(P\le\dfrac{3}{25}+\dfrac{4}{25}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{15}{25}=\dfrac{3}{5}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)

NV
17 tháng 2 2022

Đặt \(P=\dfrac{a}{b+c}+\dfrac{b}{c+d}+\dfrac{c}{a+d}+\dfrac{d}{a+b}\)

\(P=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+bd}+\dfrac{c^2}{ac+cd}+\dfrac{d^2}{ad+bd}\)

\(P\ge\dfrac{\left(a+b+c+d\right)^2}{ab+2ac+bc+2bd+cd+ad}=\dfrac{\left(a+c\right)^2+\left(b+d\right)^2+2\left(a+c\right)\left(b+d\right)}{2ac+2bd+ab+bc+cd+ad}\)

\(P\ge\dfrac{4ac+4bd+2ab+2bc+2cd+2ad}{2ac+2bd+ab+bc+cd+ad}=2\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=d\)

17 tháng 2 2022

Cám ơn thầy Lâm nhiều lắm ạ!

NV
6 tháng 4 2022

Ta có:

\(\left(b^2+c^2+1\right)\left(1+1+a^2\right)\ge\left(a+b+c\right)^2=9\)

\(\Rightarrow\dfrac{1}{b^2+c^2+1}\le\dfrac{a^2+2}{9}\)

\(\Rightarrow\dfrac{a}{b^2+c^2+1}\le\dfrac{a^3+2a}{9}\)

Tương tự: \(\dfrac{b}{c^2+a^2+1}\le\dfrac{b^3+2b}{9}\) ; \(\dfrac{c}{a^2+b^2+1}\le\dfrac{c^3+2c}{9}\)

Cộng vế:

\(VT\le\dfrac{a^3+b^3+c^3+2\left(a+b+c\right)}{9}=\dfrac{a^3+b^3+c^3+6}{9}\) (1)

Lại có:

\(\left(a^3+1+1\right)+\left(b^3+1+1\right)+\left(c^3+1+1\right)\ge3a+3b+3c\)

\(\Rightarrow a^3+b^3+c^3\ge3\Rightarrow6\le2\left(a^3+b^3+c^3\right)\) (2)

(1);(2) \(\Rightarrow VT\le\dfrac{a^3+b^3+c^3+2\left(a^3+b^3+c^3\right)}{9}=\dfrac{a^3+b^3+c^3}{3}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

6 tháng 4 2022

Cách giải của  thầy rất tự nhiên, em cám ơn thầy ạ!