K
Khách

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

29 tháng 1 2020

Ta sẽ chứng minh: \(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\)với x,y > 0.

Thật vậy: \(x+y+z\ge3\sqrt[3]{xyz}\)(bđt Cô -si)

và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt[3]{\frac{1}{abc}}\)(bđt Cô -si)

\(\Rightarrow\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\)(Dấu "="\(\Leftrightarrow x=y=z\))

Ta có: \(5a^2+2ab+2b^2=\left(2a+b\right)^2+\left(a-b\right)^2\ge\left(2a+b\right)^2\)

\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)

(Dấu "=" xảy ra khi a = b)

Tương tự ta có:\(\frac{1}{\sqrt{5b^2+2bc+2c^2}}\le\frac{1}{2b+c}\le\frac{1}{9}\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\)(Dấu "=" xảy ra khi b=c)

\(\frac{1}{\sqrt{5c^2+2ca+2a^2}}\le\frac{1}{2c+a}\le\frac{1}{9}\left(\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\)(Dấu "=" xảy ra khi c=a)

\(VT=\text{Σ}_{cyc}\frac{1}{\sqrt{5a^2+2ab+b^2}}\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)\)

\(\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{2}{3}\)

(Dấu "=" xảy ra khi \(a=b=c=\frac{3}{2}\))

30 tháng 1 2020

Ô, thanh you, bạn 2k7 sao mà giỏi thế

12 tháng 4 2020

với mọi x,y,z >0 ta có: \(x+y+z\ge3\sqrt[3]{xyz};\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt[3]{\frac{1}{xyz}}\)

\(\Rightarrow\left(x+y+z\right)\left(\frac{1}{z}+\frac{1}{y}+\frac{1}{z}\right)\ge9\)

\(\Rightarrow\frac{1}{x+y+z}\le\frac{1}{9}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

đẳng thức xảy ra khi x=y=z

ta có: \(5a^2+2ab+2b^2=\left(2a+b\right)^2+\left(a-b\right)^2\ge\left(2a+b\right)^2\)

\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)

đẳng thức xảy ra khi a=b

tương tự: \(\frac{1}{\sqrt{5b^2+2ab+2b^2}}\le\frac{1}{2b+c}\le\frac{1}{9}\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\)

đẳng thức xảy ra khi b=c

\(\frac{1}{\sqrt{5c^2+2bc+2c^2}}\le\frac{1}{2c+a}\le\frac{1}{9}\left(\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\)

đẳng thức xảy ra khi c=a

Vậy \(\frac{1}{\sqrt{5a^2+2ca+2a^2}}+\frac{1}{\sqrt{5b^2+2bc+2c^2}}+\frac{1}{\sqrt{5c^2+2ac+2a^2}}\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)\)

\(\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{2}{3}\)

đẳng thức xảy ra khi a=b=c=\(\frac{3}{2}\)

29 tháng 1 2020

Tham khảo bài của mình

11 tháng 7 2017

ÁP dụng BĐT AM-Gm  ta có: 

\(Σ\frac{a^2}{\left(ab+2\right)\left(2ab+1\right)}\ge\frac{4}{9}\cdotΣ\frac{a^2}{\left(ab+1\right)^2}\)

ĐẶt \(a=\frac{x}{y};b=\frac{y}{z};c=\frac{z}{x}\) thì cần cm

\(Σ\frac{a^2}{\left(ab+1\right)^2}=Σ\left(\frac{xz}{y\left(x+z\right)}\right)^2\ge\frac{3}{4}\)

\(Σ\left(\frac{xz}{y\left(x+z\right)}\right)^2\ge\frac{1}{3}\left(\frac{xz}{y\left(x+z\right)}\right)^2\)

Theo C-S \(Σ\frac{xz}{y\left(x+z\right)}=\frac{\left(xz\right)^2}{xyz\left(x+z\right)}\ge\frac{\left(Σxy\right)^2}{2xy\left(Σx\right)}\ge\frac{3}{2}\)

\(\frac{1}{3}\cdot\left(Σ\frac{xz}{y\left(x+z\right)}\right)^2\ge\frac{1}{3}\cdot\frac{9}{4}=\frac{3}{4}\)

Đúng hay ta có ĐPCM xyar ra khi a=b=c=1

27 tháng 11 2019

Cách : AM - GM :

\(VT=3-\left(\frac{2ab^2}{2ab^2+1}+\frac{2bc^2}{2bc^2+1}+\frac{2ca^2}{2ca^2+1}\right)\left(1\right)\)

Áp dụng BĐT AM - GM :

\(\frac{2ab^2}{2ab^2+1}+\frac{2bc^2}{2bc^2+1}+\frac{2ca^2}{2ca^2+1}=\frac{2ab^2}{ab^2+ab^2+1}+\frac{2bc^2}{bc^2+bc^2+1}+\frac{2ca^2}{ca^2+ca^2+1}\)

\(\le\frac{2ab^2}{3\sqrt[3]{a^2b^4}}+\frac{2bc^2}{3\sqrt[3]{b^2c^4}}+\frac{2ca^2}{3\sqrt[3]{c^aa^4}}=\frac{2}{3}\left(\sqrt[3]{ab^2}+\sqrt[3]{bc^2}+\sqrt[3]{ca^2}\right)\)

\(\le\frac{2}{3}\left(\frac{a+b+b}{3}+\frac{b+c+c}{3}+\frac{c+a+a}{3}\right)=\frac{2}{3}\left(a+b+c\right)=2\left(2\right)\)

Từ (1) và (2) \(\Rightarrow VT\ge3-2=1\left(đpcm\right)\)

7 tháng 3 2020

Ta sẽ chứng minh :

\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\) với x, y > 0

Thật vậy : \(x+y+z\ge3\sqrt[3]{xyz}\)( bđt Cô - si )

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt[3]{\frac{1}{abc}}\) ( bđt Cô - si )

\(\Rightarrow x+y+z\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\) ( Dấu " = " \(\Leftrightarrow x=y=z\) )

Ta có :

\(5a^2+2ab+2b^2=\left(2a+b\right)^2+\left(a-b\right)^2\ge\left(2a+b\right)^2\)

\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)

( Dấu " = " xay ra khi a=b)

Tương tự ta cũng có :

\(\frac{1}{\sqrt{5b^2+2bc+2c^2}}\le\frac{1}{2b+c}\le\frac{1}{9}\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\) ( Dấu " = " xảy ra khi b=c)

\(\frac{1}{\sqrt{5c^2+2ca+2a^2}}\le\frac{1}{2c+a}\le\frac{1}{9}\left(\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\) ( Dấu " = " xay ra khi c = a )

\(VT=\sum_{cyc}\frac{1}{\sqrt{5a^2+2ab+b^2}}\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)\)

\(\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{2}{3}\)

Dấu " = " xay ra khi \(a=b=c=\frac{2}{3}\)

Chúc bạn học tốt !!

NV
7 tháng 3 2020

\(\frac{1}{\sqrt{4a^2+2ab+b^2+a^2+b^2}}\le\frac{1}{\sqrt{4a^2+2ab+b^2+2ab}}=\frac{1}{\sqrt{\left(2a+b\right)^2}}=\frac{1}{2a+b}=\frac{1}{a+a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)

\(\Rightarrow VT\le\frac{1}{9}\left(\frac{2}{a}+\frac{1}{b}+\frac{2}{b}+\frac{1}{c}+\frac{2}{c}+\frac{1}{a}\right)=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{2}{3}\)

Dấu "=" xảy ra khi \(a=b=c=\frac{2}{3}\)

17 tháng 8 2018

Áp dụng BĐT Cauchy-SChwarz ta có:

\(VT=\frac{a^4}{a^2+2a^2bc}+\frac{b^4}{b^2+2ab^2c}+\frac{c^4}{c^2+2abc^2}\)

\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^2+b^2+c^2+2abc\left(a+b+c\right)}\)

\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^2+b^2+c^2+2\cdot\frac{\left(ab+bc+ca\right)^2}{3}}\)

\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^2+b^2+c^2+2\cdot\frac{\left(a^2+b^2+c^2\right)^2}{3}}\)

\(\ge\frac{1^2}{1+2\cdot\frac{1^2}{3}}=\frac{3}{5}=VP\)

Dấu "=" bạn tự nghiên cứu nhé :D

9 tháng 9 2018

DẤU BẰNG XẢY RA\(\Leftrightarrow a=b=c=\frac{1}{\sqrt{3}}\) CÁI NÀY LÀ ĐIỂM RƠI NHÉ.

AH
Akai Haruma
Giáo viên
31 tháng 7 2019

Lời giải:

Ta thấy:

\(\text{VT}=\frac{c^2}{2ab^2c^2+c^2}+\frac{a^2}{2bc^2a^2+a^2}+\frac{b^2}{2ca^2b^2+b^2}\)

Áp dụng BĐT Bunhiacopxky:

\(\text{VT}(2ab^2c^2+c^2+2bc^2a^2+a^2+2ca^2b^2+b^2)\geq (c+a+b)^2\)

\(\Leftrightarrow \text{VT}\geq \frac{(a+b+c)^2}{2abc(ab+bc+ac)+a^2+b^2+c^2}(*)\)

Áp dụng BĐT Am-GM:

\(3=a+b+c\geq 3\sqrt[3]{abc}\Rightarrow abc\leq 1\)

\(\Rightarrow 2abc(ab+bc+ac)\leq 2(ab+bc+ac)\)

\(\Rightarrow \frac{(a+b+c)^2}{2abc(ab+bc+ac)+a^2+b^2+c^2}\geq \frac{(a+b+c)^2}{2(ab+bc+ac)+a^2+b^2+c^2}=1(**)\)

Từ \((*); (**)\Rightarrow \text{VT}\geq 1\)

Ta có đpcm. Dấu "=" xảy ra khi $a=b=c=1$

AH
Akai Haruma
Giáo viên
31 tháng 7 2019

Cách khác bằng AM-GM:

\(\text{VT}=3-\left(\frac{2ab^2}{2ab^2+1}+\frac{2bc^2}{2bc^2+1}+\frac{2ca^2}{2ca^2+1}\right)(1)\)

Áp dụng BĐT AM-GM:

\(\frac{2ab^2}{2ab^2+1}+\frac{2bc^2}{2bc^2+1}+\frac{2ca^2}{2ca^2+1}=\frac{2ab^2}{ab^2+ab^2+1}+\frac{2bc^2}{bc^2+bc^2+1}+\frac{2ca^2}{ca^2+ca^2+1}\)

\(\leq \frac{2ab^2}{3\sqrt[3]{a^2b^4}}+\frac{2bc^2}{3\sqrt[3]{b^2c^4}}+\frac{2ca^2}{3\sqrt[3]{c^2a^4}}=\frac{2}{3}(\sqrt[3]{ab^2}+\sqrt[3]{bc^2}+\sqrt[3]{ca^2})\)

\(\leq \frac{2}{3}\left(\frac{a+b+b}{3}+\frac{b+c+c}{3}+\frac{c+a+a}{3}\right)=\frac{2}{3}(a+b+c)=2(2)\)

Từ \((1);(2)\Rightarrow \text{VT}\geq 3-2=1\) (đpcm)

29 tháng 5 2018

Ta có: \(5a^2+2ab+2b^2=4a^2+2ab+b^2+\left(a^2+b^2\right)\ge4a^2+2ab+b^2+2ab=\left(2a+b\right)^2\)

\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\)

Lại có: \(\frac{1}{2a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{9}\left(\frac{2}{a}+\frac{1}{b}\right)\)

Tương tự cộng lại ta có: \(VT\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Theo BĐT Bunhiacopxki ta có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\le3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=3\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\sqrt{3}\)

\(\Rightarrow VT\le\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}\)

Dấu = xảy ra khi \(a=b=c=\sqrt{3}\)

NV
4 tháng 3 2019

\(5a^2+2ab+2b^2=\left(2a+b\right)^2+\left(a-b\right)^2\ge\left(2a+b\right)^2\)

\(\Rightarrow\dfrac{1}{\sqrt{5a^2+2ab+2b^2}}\le\dfrac{1}{\sqrt{\left(2a+b\right)^2}}=\dfrac{1}{a+a+b}\le\dfrac{1}{9}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}\right)\)

Tương tự ta có: \(\dfrac{1}{\sqrt{5b^2+2bc+2c^2}}\le\dfrac{1}{9}\left(\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\dfrac{1}{\sqrt{5c^2+2ac+a^2}}\le\dfrac{1}{9}\left(\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a}\right)\)

Cộng vế với vế:

\(\dfrac{1}{\sqrt{5a^2+2ab+b^2}}+\dfrac{1}{\sqrt{5b^2+2bc+c^2}}+\dfrac{1}{\sqrt{5c^2+2ac+a^2}}\le\dfrac{1}{9}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)\le\dfrac{2}{3}\)

Dấu "=" khi \(a=b=c=\dfrac{3}{2}\)