K
Khách

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

9 tháng 8 2017

Đặt \(T=\left(a+b\right)\left(b+c\right)\left(c+a\right)>0\)

\(BDT\Leftrightarrow\dfrac{a^2+bc}{b+c}+\dfrac{b^2+ca}{c+a}+\dfrac{c^2+ab}{a+b}\ge a+b+c\)

\(\Leftrightarrow\dfrac{a^2+bc}{b+c}-a+\dfrac{b^2+ca}{c+a}-b+\dfrac{c^2+ab}{a+b}-c\ge0\)

\(\Leftrightarrow\dfrac{a^2+bc-ab-ac}{b+c}+\dfrac{b^2+ac-ab-bc}{a+c}+\dfrac{c^2+ab-ac-bc}{a+b}\ge0\)

\(\Leftrightarrow\dfrac{\left(a-b\right)\left(a-c\right)}{b+c}+\dfrac{\left(b-a\right)\left(b-c\right)}{a+c}+\dfrac{\left(c-a\right)\left(c-b\right)}{a+b}\ge0\)

\(\Leftrightarrow\dfrac{\left(a^2-b^2\right)\left(a^2-c^2\right)+\left(b^2-a^2\right)\left(b^2-c^2\right)+\left(c^2-a^2\right)\left(c^2-b^2\right)}{T}\ge0\)

\(\Leftrightarrow\dfrac{a^4+b^4+c^4-b^2c^2-c^2a^2-a^2b^2}{T}\ge0\)

\(\Leftrightarrow\dfrac{\left(a^2-b^2\right)^2+\left(b^2-c^2\right)^2+\left(c^2-a^2\right)^2}{2T}\ge0\)

Xảy ra khi \(a=b=c\)

10 tháng 8 2017

\(BĐT\Leftrightarrow\sum\left(\dfrac{1}{a}-\dfrac{b+c}{a^2+bc}\right)\ge0\)

\(\Leftrightarrow\sum\dfrac{\left(a-b\right)\left(a-c\right)}{a\left(a^2+bc\right)}\ge0\)

Giả sử \(a\ge b\ge c\)thì

\(\dfrac{\left(a-b\right)\left(a-c\right)}{a\left(a^2+bc\right)}\ge0\).vậy nên chỉ cần chứng minh

\(\dfrac{\left(b-c\right)\left(b-a\right)}{b\left(b^2+ac\right)}+\dfrac{\left(c-a\right)\left(c-b\right)}{c\left(c^2+ab\right)}\ge0\)

\(\Leftrightarrow\left(b-c\right)\left[\dfrac{b-a}{b\left(b^2+ac\right)}+\dfrac{a-c}{c\left(c^2+ab\right)}\right]\ge0\)

\(\Leftrightarrow\left(b-c\right)\left[\left(b-a\right)\left(c^3+abc\right)+\left(a-c\right)\left(b^3+abc\right)\right]\ge0\)

\(\Leftrightarrow\left(b-c\right)^2\left(b+c\right)\left(ab+ac-bc\right)\ge0\)( đúng vì \(a\ge b\ge c\))

Vậy BĐT được chứng minh.

Dấu = xảy ra khi a=b=c

6 tháng 3 2021

\(\dfrac{a}{\sqrt{a^2+1}}=\dfrac{a}{\sqrt{a^2+ab+ac+bc}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{a}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)=\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\) Chứng minh tương tự ta được:

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

\(\Rightarrow\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{b+a}+\dfrac{b}{b+c}+\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)=\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{1}{2}\left(1+1+1\right)=\dfrac{3}{2}\) Dấu = xảy ra \(\Leftrightarrow a=b=c=\dfrac{1}{\sqrt{3}}\)

NV
6 tháng 3 2021

\(\dfrac{a}{\sqrt{a^2+1}}=\dfrac{a}{\sqrt{a^2+ab+bc+ca}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)

Tương tự: \(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)\) ; \(\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{b+c}\right)\)

Cộng vế:

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

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

AH
Akai Haruma
Giáo viên
1 tháng 3 2022

Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{b+c}{a^2+bc}=\frac{(b+c)^2}{(b+c)(a^2+bc)}=\frac{(b+c)^2}{b(a^2+c^2)+c(a^2+b^2)}\leq \frac{c^2}{b(a^2+c^2)}+\frac{b^2}{c(a^2+b^2)}\)

Tương tự với các phân thức còn lại:

$\frac{c+a}{b^2+ca}\leq \frac{c^2}{b(a^2+c^2)}+\frac{a^2}{c(a^2+b^2)}$

$\frac{a+b}{c^2+ab}\leq \frac{a^2}{b(a^2+c^2)}+\frac{b^2}{c(a^2+b^2)}$

Cộng theo vế và thu gọn suy ra:

$\text{VT}\leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ (đpcm)

AH
Akai Haruma
Giáo viên
1 tháng 3 2022

Dấu "=" xảy ra khi a=b=c

10 tháng 8 2017

\(BĐT\Leftrightarrow\sum\dfrac{2bc}{1+a^2}\le\dfrac{3}{2}\Leftrightarrow\sum\dfrac{-2bc}{2a^2+b^2+c^2}\ge-\dfrac{3}{2}\)

\(\Leftrightarrow\sum\dfrac{2a^2+\left(b-c\right)^2}{2a^2+b^2+c^2}\ge\dfrac{3}{2}\)

ÁP dụng BĐT cauchy-schwarz:

\(\sum\dfrac{2a^2}{2a^2+b^2+c^2}\ge\dfrac{2\left(a+b+c\right)^2}{4\left(a^2+b^2+c^2\right)}=\dfrac{\left(a+b+c\right)^2}{2\left(a^2+b^2+c^2\right)}\)

\(\sum\dfrac{\left(b-c\right)^2}{2a^2+b^2+c^2}=\dfrac{\left(b-c\right)^2}{2a^2+b^2+c^2}+\dfrac{\left(a-b\right)^2}{2c^2+a^2+b^2}+\dfrac{\left(a-c\right)^2}{2b^2+a^2+c^2}\ge\dfrac{4\left(a-c\right)^2}{4\left(a^2+b^2+c^2\right)}=\dfrac{\left(a-c\right)^2}{a^2+b^2+c^2}\)

( Lưu ý : \(\left(c-a\right)^2=\left(a-c\right)^2\)) (1)

Do vậy cần chứng minh \(\dfrac{\left(a+b+c\right)^2+2\left(a-c\right)^2}{2\left(a^2+b^2+c^2\right)}\ge\dfrac{3}{2}\)

\(\Leftrightarrow2\left(a+b+c\right)^2+4\left(a-c\right)^2\ge6\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow ab+bc-ac-b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)\left(b-c\right)\ge0\) (*)

(*) không phải luôn đúng, tuy nhiên ta có thể ép cho nó đúng .

bằng cách đáng giá tương tự BĐT (1) :

\(\left\{{}\begin{matrix}\dfrac{\left(b-a\right)^2}{2c^2+a^2+b^2}+\dfrac{\left(b-c\right)^2}{2a^2+b^2+c^2}+\dfrac{\left(c-a\right)^2}{2b^2+a^2+c^2}\ge\dfrac{\left(b-a\right)^2}{a^2+b^2+c^2}\\\dfrac{\left(a-b\right)^2}{2c^2+a^2+b^2}+\dfrac{\left(c-b\right)^2}{2a^2+b^2+c^2}+\dfrac{\left(c-a\right)^2}{2b^2+a^2+c^2}\ge\dfrac{\left(c-b\right)^2}{a^2+b^2+c^2}\end{matrix}\right.\)

ta thu được BĐT cần chứng minh tương đương \(\left\{{}\begin{matrix}\left(b-c\right)\left(c-a\right)\ge0\left(3\right)\\\left(c-a\right)\left(a-b\right)\ge0\left(4\right)\end{matrix}\right.\)

Dễ thấy \(\left(a-b\right)\left(b-c\right).\left(b-c\right)\left(c-a\right).\left(c-a\right)\left(a-b\right)=\left[\left(a-b\right)\left(b-c\right)\left(c-a\right)\right]^2\ge0\)

tích của chúng là 1 số không âm nên có ít nhất 1 số không âm .Chứng tỏ có ít nhất 1 BĐT đúng

Do đó ta có đpcm

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

2 tháng 3 2021

Áp dụng BĐT BSC:

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

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

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

\(\ge a+b+c-\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)

Đẳng thức xảy ra khi \(a=b=c\)

2 tháng 3 2021

4ab ≤ (a + b)2 ⇒ \(\dfrac{4ab}{a+b}\le a+b\)

Tương tự \(\dfrac{4ac}{a+c}\le a+c\) ; \(\dfrac{4bc}{b+c}\le b+c\)

⇒ Cộng lại vế với vế :

4VT ≤ 2 (a+b+c) ⇒ VT ≤ \(\dfrac{a+b+c}{2}\)

NV
1 tháng 5 2021

a.

Ta có: \(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge2\sqrt{\dfrac{a^2\left(b+c\right)}{4\left(b+c\right)}}=a\)

Tương tự: \(\dfrac{b^2}{c+a}+\dfrac{c+a}{4}\ge b\) ; \(\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge c\)

Cộng vế:

\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{a+b+c}{2}\ge a+b+c\)

\(\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

NV
1 tháng 5 2021

b.

Ta có:

\(a^2+bc\ge2\sqrt{a^2bc}=2\sqrt{ab.ac}\Rightarrow\dfrac{1}{a^2+bc}\le\dfrac{1}{2\sqrt{ab.ac}}\le\dfrac{1}{4}\left(\dfrac{1}{ab}+\dfrac{1}{ac}\right)\)

Tương tự: \(\dfrac{1}{b^2+ac}\le\dfrac{1}{4}\left(\dfrac{1}{ab}+\dfrac{1}{bc}\right)\) ; \(\dfrac{1}{c^2+ab}\le\dfrac{1}{4}\left(\dfrac{1}{ac}+\dfrac{1}{bc}\right)\)

Cộng vế với vế:

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

Dấu "=" xảy ra khi \(a=b=c\)

13 tháng 6 2021

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

Áp dụng các bđt sau:Với x;y;z>0 có: \(\dfrac{1}{x+y+z}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) và \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\) 

Có \(\dfrac{1}{a+3b+2c}=\dfrac{1}{\left(a+b\right)+\left(b+c\right)+\left(b+c\right)}\le\dfrac{1}{9}\left(\dfrac{1}{a+b}+\dfrac{2}{b+c}\right)\)\(\le\dfrac{1}{9}.\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{2}{c}\right)=\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{3}{b}+\dfrac{2}{c}\right)\)

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

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

Cộng vế với vế => \(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{36}.6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)

Dấu = xảy ra khi a=b=c=3

13 tháng 6 2021

Có \(a+b=2\Leftrightarrow2\ge2\sqrt{ab}\Leftrightarrow ab\le1\)

\(E=\left(3a^2+2b\right)\left(3b^2+2a\right)+5a^2b+5ab^2+2ab\)

\(=9a^2b^2+6\left(a^3+b^3\right)+4ab+5ab\left(a+b\right)+20ab\)

\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+4ab+5ab\left(a+b\right)+20ab\)

\(=9a^2b^2+48-18ab.2+4ab+5.2.ab+20ab\)

\(=9a^2b^2-2ab+48\)

Đặt \(f\left(ab\right)=9a^2b^2-2ab+48;ab\le1\), đỉnh \(I\left(\dfrac{1}{9};\dfrac{431}{9}\right)\)

Hàm đồng biến trên khoảng \(\left[\dfrac{1}{9};1\right]\backslash\left\{\dfrac{1}{9}\right\}\)

 \(\Rightarrow f\left(ab\right)_{max}=55\Leftrightarrow ab=1\)

\(\Rightarrow E_{max}=55\Leftrightarrow a=b=1\)

Vậy...

31 tháng 3 2017

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

\(\Rightarrow ab+bc+ca=abc\)

Xét \(\dfrac{a^2}{a+bc}+\dfrac{b^2}{b+ca}+\dfrac{c^2}{c+ab}\)

\(\Leftrightarrow\dfrac{a^3}{a^2+abc}+\dfrac{b^3}{b^2+abc}+\dfrac{c^3}{c^2+abc}\)

\(\Leftrightarrow\dfrac{a^3}{a^2+ab+bc+ca}+\dfrac{b^3}{b^2+ab+bc+ca}+\dfrac{c^3}{c^2+ab+bc+ca}\)

\(\Leftrightarrow\dfrac{a^3}{a\left(a+b\right)+c\left(a+b\right)}+\dfrac{b^3}{b\left(a+b\right)+c\left(a+b\right)}+\dfrac{c^3}{c\left(b+c\right)+a\left(b+c\right)}\)

\(\Leftrightarrow\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{c^3}{\left(b+c\right)\left(c+a\right)}\)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3a}{4}\\\dfrac{b^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{a+b}{8}+\dfrac{b+c}{8}\ge3\sqrt[3]{\dfrac{b^3}{64}}=\dfrac{3b}{4}\\\dfrac{b^3}{\left(b+c\right)\left(c+a\right)}+\dfrac{b+c}{8}+\dfrac{c+a}{8}\ge3\sqrt[3]{\dfrac{b^3}{64}}=\dfrac{3b}{4}\end{matrix}\right.\)

\(\Rightarrow\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{c^3}{\left(b+c\right)\left(c+a\right)}+\dfrac{4\left(a+b+c\right)}{8}\ge\dfrac{3\left(a+b+c\right)}{4}\)

\(\Rightarrow\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{c^3}{\left(b+c\right)\left(c+a\right)}+\dfrac{a+b+c}{2}\ge\dfrac{3\left(a+b+c\right)}{4}\)

\(\Rightarrow\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{c^3}{\left(b+c\right)\left(c+a\right)}\ge\dfrac{3\left(a+b+c\right)}{4}-\dfrac{a+b+c}{2}\)

\(\Rightarrow\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{c^3}{\left(b+c\right)\left(c+a\right)}\ge\dfrac{a+b+c}{4}\)

\(\Leftrightarrow\dfrac{a^2}{a+bc}+\dfrac{b^2}{b+ca}+\dfrac{c^2}{c+ab}\ge\dfrac{a+b+c}{4}\) ( đpcm )

Dấu " = " xảy ra khi \(a=b=c=3\)

p/s: bài này em nhớ em đã giải cho anh ròi mà ta =))

3 tháng 4 2017

đài thế cách tui ngắn hơn nhiều

NV
8 tháng 4 2021

\(\Leftrightarrow\left(1+ab+bc+ca\right)\left(\dfrac{1}{\left(a+b\right)\left(a+c\right)}+\dfrac{1}{\left(a+b\right)\left(b+c\right)}+\dfrac{1}{\left(a+c\right)\left(b+c\right)}\right)\le\dfrac{ab+bc+ca}{abc}\)

\(\Leftrightarrow\dfrac{2\left(1+ab+bc+ca\right)\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{ab+bc+ca}{abc}\)

\(\Leftrightarrow\dfrac{2\left(1+ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{ab+bc+ca}{abc}\)

Áp dụng BĐT quen thuộc:

\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\dfrac{8}{9}\left(ab+bc+ca\right)\left(a+b+c\right)=\dfrac{8}{9}\left(ab+bc+ca\right)\)

\(\Rightarrow\dfrac{2\left(1+ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{9\left(1+ab+bc+ca\right)}{4\left(ab+bc+ca\right)}\)

Ta chỉ cần chứng minh:

\(\dfrac{9\left(1+ab+bc+ca\right)}{4\left(ab+bc+ca\right)}\le\dfrac{ab+bc+ca}{abc}\)

\(\Leftrightarrow4\left(ab+bc+ca\right)^2\ge9abc+9abc\left(ab+bc+ca\right)\)

Do \(3\left(ab+bc+ca\right)^2\ge9abc\left(a+b+c\right)=9abc\)

Nên ta chỉ cần chứng minh:

\(\left(ab+bc+ca\right)^2\ge9abc\left(ab+bc+ca\right)\)

\(\Leftrightarrow ab+bc+ca\ge9abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge9\)

Hiển nhiên đúng do \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}=9\)