cho a+b+c=2;chứng minh rằng (2-c)(b-c)/2a+bc+(2-a)(c-a)/2b+ca+(2-b)(a-b)/2c+ab lớn hơn hoặc bằng 0
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DK
29 tháng 6 2016
a) Ta có: \(a+b+c=0\)
\(\Rightarrow a^2+b^2+c^2+2ab+2ac+2bc=0\)
\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)
\(\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)\)
\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left[a^2b^2+b^2c^2+c^2a^2+2abc\left(b+a+c\right)\right]\)
\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(\Rightarrow a^4+b^4+c^4=4\left(a^2b^2+b^2c^2+c^2a^2\right)-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
DK
29 tháng 6 2016
b) Ta có: \(a+b+c=0\)
\(\Rightarrow2abc\left(a+b+c\right)=0\)
\(\Rightarrow2a^2bc+2ab^2c+2abc^2=0\)
Ta lại có:
\(a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)^2\)(chứng minh câu a)
\(\Rightarrow a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2+4a^2bc+4ab^2c+4abc^2\)
\(\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)\)
\(\Rightarrow a^4+b^4+c^4=2\left(ab+bc+ca\right)^2\)
QK
8 tháng 4 2017
a2b+ab2-2abc +b2c+bc2-2abc+ac2+a2c-2abc
=b(a2-2ac+c2) +a(b2-2bc+c2)+c (a2-2ab+b2)
= b(a-c)2+a(b-c)2+c(a-b)2 vì a,b,c là độ dài ba cạnh tam giác=) a,b,c>0
b(a-c)2\(\ge0\) \(\forall a,b,c\)
a(b-c)2\(\ge0\)\(\forall a,b,c\)
c(a-b)2\(\ge0\forall a,b,c\)
18 tháng 9 2018
\(\sqrt{\dfrac{a+b}{c+ab}}+\sqrt{\dfrac{b+c}{a+bc}}+\sqrt{\dfrac{c+a}{b+ac}}\)
N
30 tháng 9 2017
Bài này có xuất hiện rồi ,you vào mục tìm kiếm là thấy liền.
Lời giải vắn tắt:
\(A=\sum\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}=\sum\dfrac{ab+2c^2}{\sqrt{\left(ab+2c^2\right)\left(1+ab-c^2\right)}}\ge\sum\dfrac{2\left(ab+2c^2\right)}{1+2ab+c^2}=\sum\dfrac{2\left(ab+2c^2\right)}{\left(a+b\right)^2+2c^2}\ge\sum\dfrac{2\left(ab+2c^2\right)}{2\left(a^2+b^2\right)+2c^2}=\sum\left(ab+2c^2\right)=ab+bc+ca+2\)
( thay \(a^2+b^2+c^2=1\))
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...
\(\frac{\left(2-c\right)\left(b-c\right)}{2a+bc}=\frac{\left(a+b\right)\left(b-c\right)}{a\left(a+b+c\right)+bc}=\frac{\left(a+b\right)\left(b-c\right)}{\left(a+b\right)\left(c+a\right)}=\frac{b-c}{c+a}=\frac{b}{c+a}-\frac{c}{c+a}\)
Tương tự, ta có: \(\frac{\left(2-a\right)\left(c-a\right)}{2b+ca}=\frac{c}{a+b}-\frac{a}{a+b};\frac{\left(2-b\right)\left(a-b\right)}{2c+ab}=\frac{a}{b+c}-\frac{b}{b+c}\)
\(\Rightarrow\)\(VT=\left(\frac{a}{b+c}-\frac{a}{a+b}\right)+\left(\frac{b}{c+a}-\frac{b}{b+c}\right)+\left(\frac{c}{a+b}-\frac{c}{c+a}\right)\)
\(=\frac{a\left(a-c\right)}{\left(a+b\right)\left(b+c\right)}+\frac{b\left(b-a\right)}{\left(b+c\right)\left(c+a\right)}+\frac{c\left(c-b\right)}{\left(c+a\right)\left(a+b\right)}\)
\(=\frac{a\left(a-c\right)\left(c+a\right)+b\left(b-a\right)\left(a+b\right)+c\left(c-b\right)\left(b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\frac{\left(a^3+b^3+c^3\right)-\left(a^2b+b^2c+c^2a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge\frac{\left(a^3+b^3+c^3\right)-\left(a^3+b^3+c^3\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{2}{3}\)
cái bđt \(a^3+b^3+c^3\ge a^2b+b^2c+c^2a\) cô Chi có làm r ib mk gửi link