Cho các số dương a,b,c. CMR:
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{4c}{a+b}>2\)
@Lightning Farron
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4 tháng 5 2018
Sửa \(\dfrac{1}{3}\rightarrow3\)
Từ \(a+b+c+ab+bc+ca=6abc\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=6\)
Ta có: \(\dfrac{1}{a^2}+1\ge\dfrac{2}{a};\dfrac{1}{b^2}+1\ge\dfrac{2}{b};\dfrac{1}{c^2}+1\ge\dfrac{2}{c}\)
Và \(\dfrac{1}{a^2}+\dfrac{1}{b^2}\ge\dfrac{2}{ab};\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{2}{bc};\dfrac{1}{c^2}+\dfrac{1}{a^2}\ge\dfrac{2}{ac}\)
Cộng theo vế các BĐT trên ta có:
\(3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+1\right)\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)
\(\Leftrightarrow3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+1\right)\ge12\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+1\ge4\)\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge3\)
\("="\Leftrightarrow a=b=c=1\)
3 tháng 3 2018
\(A=\frac{a}{b+c}+\frac{b}{c+a}+\frac{4c}{a+b}\)
\(A>\frac{2a}{a+b+c}+\frac{2b}{c+a+b}+\frac{2c}{a+b+c}\)
\(A>\frac{2\left(a+b+c\right)}{a+b+c}>2\Rightarrow dpcm\)
13 tháng 7 2020
\(\Sigma_{sym}a^4b^4\ge\frac{\left(\Sigma_{sym}a^2b^2\right)^2}{3}\ge\frac{\left(\Sigma_{sym}ab\right)^4}{27}\ge\frac{a^2b^2c^2\left(a+b+c\right)^2}{3}=3a^4b^4c^4\)
13 tháng 7 2020
\(\Sigma\frac{a^5}{bc^2}\ge\frac{\left(a^3+b^3+c^3\right)^2}{abc\left(a+b+c\right)}\ge\frac{\left(a^2+b^2+c^2\right)^4}{abc\left(a+b+c\right)^3}\ge\frac{\left(a+b+c\right)^6\left(a^2+b^2+c^2\right)}{27abc\left(a+b+c\right)^3}\)
\(\ge\frac{\left(3\sqrt[3]{abc}\right)^3\left(a^2+b^2+c^2\right)}{27abc}=a^2+b^2+c^2\)
CM
1
8 tháng 5 2018
Áp dụng BĐT Cauchy Sshwarz, ta có:
\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\)
Mà a+b+c>2
\(\Rightarrow VT>1\) (đpcm)
TN
2 tháng 7 2016
Bài 1:
Đặt \(a^2=x;b^2=y;c^2=z\)
Ta có:\(\sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}}\le\frac{3}{\sqrt{2}}\)
Áp dụng BĐT cô si ta có:
\(\sqrt{\frac{x}{x+y}}=\frac{1}{\sqrt{2}}\sqrt{\frac{4x\left(x+y+z\right)}{3\left(x+y\right)\left(x+z\right)}\frac{3\left(x+z\right)}{2\left(x+y+z\right)}}\)
\(\le\frac{1}{2\sqrt{2}}\left[\frac{4x\left(x+y+z\right)}{3\left(x+y\right)\left(x+z\right)}+\frac{3\left(x+z\right)}{2\left(x+y+z\right)}\right]\)
Tương tự với \(\sqrt{\frac{y}{y+z}}\)và \(\sqrt{\frac{z}{z+x}}\)
Cộng lại ta được:
\(\frac{\sqrt{2}}{3}\left[\frac{x\left(x+y+z\right)}{\left(x+y\right)\left(x+z\right)}+\frac{y\left(x+y+z\right)}{\left(y+z\right)\left(y+x\right)}+\frac{z\left(x+y+z\right)}{\left(z+x\right)\left(z+y\right)}\right]+\frac{3}{2\sqrt{2}}\le\frac{3}{2\sqrt{2}}\)
Sau đó bình phương hai vế rồi
\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge8xyz\)đẳng thức đúng
Vậy...
Bài 2:
Trước hết ta chứng minh bất đẳng thức sau:
\(\frac{a}{4a+4b+c}+\frac{b}{4b+4c+a}+\frac{c}{4c+4a+b}\le\frac{1}{3}\)
Nhân cả hai vế bđt với 4(a+b+c)4(a+b+c) rồi thu gọn ta được bđt sau:
\(\frac{4a\left(a+b+c\right)}{4a+4b+c}+\frac{4b\left(a+b+c\right)}{4b+4c+a}+\frac{4c\left(a+b+c\right)}{4c+4a+b}\)\(\le\frac{4}{3}\left(a+b+c\right)\)
\(\left[\frac{4a\left(a+b+c\right)}{4a+4b+}-a\right]+\left[\frac{4b\left(a+b+c\right)}{4b+4c+a}-b\right]+\left[\frac{4c\left(a+b+c\right)}{4c+4a+b}-c\right]\le\frac{a+b+c}{3}\)
\(\frac{ca}{4a+4b+c}+\frac{ab}{4b+4c+a}+\frac{bc}{4c+4a+b}\le\frac{a+b+c}{9}\)
Áp dụng bđt cauchy-Schwarz ta có \(\frac{ca}{4a+4b+c}=\frac{ca}{\left(2b+c\right)+2\left(2a+b\right)}\)\(\le\frac{ca}{9}\left(\frac{1}{2b+c}+\frac{2}{2a+b}\right)\)
Từ đó ta có:
\(\text{∑}\frac{ca}{4a+4b+c}\le\frac{1}{9}\text{∑}\left(\frac{ca}{2b+c}+\frac{2ca}{2a+b}\right)\)\(=\frac{1}{9}\left(\text{ ∑}\frac{ca}{2b+c}+\text{ ∑}\frac{2ca}{2a+b}\right)\)\(=\frac{1}{9}\left(\text{ ∑}\frac{ca}{2b+c}+\text{ ∑}\frac{2ab}{2b+c}\right)=\frac{a+b+c}{9}\)
Đặt VT=A rồi áp dụng bđt cauchy-Schwarz cho VT ta có
\(T^2\le3\left(\frac{a}{4a+4b+c}+\frac{b}{4b+4c+a}+\frac{c}{4c+4a+b}\right)\)\(\le3\cdot\frac{1}{3}=1\Leftrightarrow T\le1\)
Dấu = xảy ra khi a=b=c
c bạn tự làm nhé mình mệt rồi :D
25 tháng 2 2018
Cauchy-Schwarz dạng ENgel:
\(P=\dfrac{a}{2a+b}+\dfrac{b}{2b+c}+\dfrac{c}{2c+a}\)
\(=\dfrac{1}{2}\cdot3-\left(\dfrac{b}{4a+2b}+\dfrac{c}{4b+2c}+\dfrac{a}{4c+2a}\right)\)
\(=\dfrac{3}{2}-\left(\dfrac{b^2}{4ab+2b^2}+\dfrac{c^2}{4bc+2c^2}+\dfrac{a^2}{4ac+2a^2}\right)\)
\(\le\dfrac{3}{2}-\dfrac{\left(a+b+c\right)^2}{2\left(a^2+b^2+c^2+2ab+2bc+2ca\right)}\)
\(=\dfrac{3}{2}-\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)^2}=1\)
\("="\Leftrightarrow a=b=c=\dfrac{1}{3}\)
NB
0
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{4c}{a+b}\ge2\)
\(VT:\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{4c}{a+b}\)
\(=\dfrac{a}{b+c}+1+\dfrac{b}{c+a}+1+\dfrac{4c}{a+b}+4-6\\ =\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}+\dfrac{4a+4b+4c}{a+b}-6\\ =\left(a+b+c\right)\cdot\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{4}{a+b}\right)-6\)
Áp dụng bđt Caychuy - Schwarz :
\ \(\left(a+b+c\right)\cdot\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{4}{a+b}\right)-6\\ \ge\left(a+b+c\right)\cdot\dfrac{\left(1+1+2\right)^2}{2a+2b+2c}-6\\ \ge\dfrac{16}{2}-6=2\)
Dấu = xảy ra khi \(a=b=c\)
Cách khác:
Ta có: \(\left(a-b\right)^2+4c^2\ge0\Leftrightarrow a^2-2ab+b^2+4c^2\ge0\)
\(\Leftrightarrow a^2+b^2+4c^2+2ab+4bc+4ac-4ab-4bc-4ca\ge0\)\(\Leftrightarrow\left(a+b+2c\right)^2-4\left(ab+bc+ca\right)\ge0\)
\(\Leftrightarrow\left(a+b+2c\right)^2\ge4\left(ab+bc+ca\right)\)
Theo BĐT cauchy-schwarz ta có:
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{4c}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ba}+\dfrac{4c^2}{ac+ab}\ge\)\(\ge\dfrac{\left(a+b+2c\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{4\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=2\Rightarrowđpcm\)