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15 tháng 7 2018

Áp dụng BĐT sau: \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) 

\(\Rightarrow\frac{1}{2a+b+c}\le\frac{1}{4}\left(\frac{1}{2a}+\frac{1}{b+c}\right)\). Lại có \(\frac{1}{b+c}\le\frac{1}{4b}+\frac{1}{4c}\)

\(\Rightarrow\frac{1}{2a+b+c}\le\frac{1}{4}\left(\frac{1}{2a}+\frac{1}{4b}+\frac{1}{4c}\right)\)

Tương tự: \(\frac{1}{a+2b+c}\le\frac{1}{4}\left(\frac{1}{4a}+\frac{1}{2b}+\frac{1}{4c}\right);\frac{1}{a+b+2c}\le\frac{1}{4}\left(\frac{1}{4a}+\frac{1}{4b}+\frac{1}{2c}\right)\)

Cộng 3 BĐT trên theo vế, ta được:

\(\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Thay \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=4\)\(\Rightarrow\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\le1\)(đpcm).

Dấu "=" xảy ra <=> \(a=b=c=\frac{3}{4}.\)

24 tháng 12 2015

CM BĐT : \(\frac{1}{a+b+c+d}\le\frac{1}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\) 

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge\frac{\left(1+1+1+1\right)^2}{a+b+c+d}=\frac{16}{a+b+c+d}\)

=> \(\frac{1}{a+b+c+d}\le\frac{1}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\)

ÁP dụng BĐT : \(\frac{1}{a+a+b+c}+\frac{1}{a+b+b+c}+\frac{1}{a+b+c+c}\le\frac{1}{16}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac{1}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)+\frac{1}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)

                               \(=\frac{1}{16}4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{16}\cdot4\cdot4=1\)

Dấu '' = '' xảy ra khi a = b= c = 3/4 

2 tháng 12 2020

Đặt \(\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)=\left(x,y,z\right)\)

\(x+y+z\ge\frac{x^2+2xy}{2x+y}+\frac{y^2+2yz}{2y+z}+\frac{z^2+2zx}{2z+x}\)

\(\Leftrightarrow x+y+z\ge\frac{3xy}{2x+y}+\frac{3yz}{2y+z}+\frac{3zx}{2z+x}\)

\(\frac{3xy}{2x+y}\le\frac{3}{9}xy\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{3}\left(x+2y\right)\)

\(\Rightarrow\Sigma_{cyc}\frac{3xy}{2x+y}\le\frac{1}{3}\left[\left(x+2y\right)+\left(y+2z\right)+\left(z+2x\right)\right]=x+y+z\)

Dấu "=" xảy ra khi x=y=z

28 tháng 4 2019

Từ \(ab+bc+ca=5abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=5\)

Áp dụng BĐT Bu-nhi-a-cốp-xki ta có :

\(\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\left(a+a+b+b+c\right)\ge\left(1+1+1+1+1\right)^2\)

\(\Rightarrow\frac{2}{a}+\frac{2}{b}+\frac{1}{c}\ge\frac{25}{2a+2b+c}\)

Tương tự ta có :

\(\frac{2}{b}+\frac{2}{c}+\frac{1}{a}\ge\frac{25}{2b+2c+a}\)

\(\frac{2}{a}+\frac{1}{b}+\frac{2}{c}\ge\frac{25}{2a+b+2c}\)

Cộng từng vế BĐT ta thu được :

\(\frac{5}{a}+\frac{5}{b}+\frac{5}{c}\ge25P\)

\(\Leftrightarrow P\le\frac{5\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}{25}=1\)

Vậy BĐT đã được chứng minh . Dấu \("="\) xảy ra khi \(a=b=c=\frac{3}{5}\)

NV
19 tháng 3 2019

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

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

Cộng vế với vế:

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

Dấu "=" xảy ra khi \(a=b=c\)

1 tháng 4 2017

Bài 1:

\(BDT\Leftrightarrow\sqrt{\frac{3}{a+2b}}+\sqrt{\frac{3}{b+2c}}+\sqrt{\frac{3}{c+2a}}\le\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\)

\(\Leftrightarrow\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\ge\sqrt{3}\left(\frac{1}{\sqrt{a+2b}}+\frac{1}{\sqrt{b+2c}}+\frac{1}{\sqrt{c+2a}}\right)\)

Áp dụng BĐT Cauchy-Schwarz và BĐT AM-GM ta có: 

\(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{b}}\ge\frac{9}{\sqrt{a}+\sqrt{2}\cdot\sqrt{2b}}\ge\frac{9}{\sqrt{\left(1+2\right)\left(a+2b\right)}}=\frac{3\sqrt{3}}{\sqrt{a+2b}}\)

Tương tự cho 2 BĐT còn lại ta cũng có: 

\(\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}+\frac{1}{\sqrt{c}}\ge\frac{3\sqrt{3}}{\sqrt{b+2c}};\frac{1}{\sqrt{c}}+\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{a}}\ge\frac{3\sqrt{3}}{\sqrt{c+2a}}\)

Cộng theo vế 3 BĐT trên ta có: 

\(3\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)\ge3\sqrt{3}\left(\frac{1}{\sqrt{a+2b}}+\frac{1}{\sqrt{b+2c}}+\frac{1}{\sqrt{c+2a}}\right)\)

\(\Leftrightarrow\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\ge\sqrt{3}\left(\frac{1}{\sqrt{a+2b}}+\frac{1}{\sqrt{b+2c}}+\frac{1}{\sqrt{c+2a}}\right)\)

Đẳng thức xảy ra khi \(a=b=c\)

Bài 2: làm mãi ko ra hình như đề sai, thử a=1/2;b=4;c=1/2

1 tháng 4 2017

Bài 2/

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

\(=\frac{b^2c^2}{a^2b^2c+a^2c^2b}+\frac{c^2a^2}{b^2c^2a+b^2a^2c}+\frac{a^2b^2}{c^2a^2b+c^2b^2a}\)

\(=\frac{b^2c^2}{ab+ac}+\frac{c^2a^2}{bc+ba}+\frac{a^2b^2}{ca+cb}\)

\(\ge\frac{\left(bc+ca+ab\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\)

\(\ge\frac{3\sqrt[3]{ab.bc.ca}}{2}=\frac{3}{2}\)

Dấu =  xảy ra khi \(a=b=c=1\)

NV
1 tháng 8 2020

\(1-\frac{a}{a+1}\ge\frac{2b}{b+1}+\frac{3c}{c+1}\Leftrightarrow\frac{1}{a+1}\ge\frac{b}{b+1}+\frac{b}{b+1}+\frac{c}{c+1}+\frac{c}{c+1}+\frac{c}{c+1}\ge5\sqrt[5]{\frac{b^2c^3}{\left(b+1\right)^2\left(c+1\right)^3}}\)

Tương tự:

\(\frac{1}{b+1}\ge\frac{a}{a+1}+\frac{b}{b+1}+3.\frac{c}{c+1}\ge5\sqrt[5]{\frac{abc^3}{\left(a+1\right)\left(b+1\right)\left(c+1\right)^3}}\)

\(\Leftrightarrow\frac{1}{\left(b+1\right)^2}\ge25\sqrt[5]{\frac{a^2b^2c^6}{\left(a+1\right)^2\left(b+1\right)^2\left(c+1\right)^6}}\)

\(\frac{1}{c+1}\ge\frac{a}{a+1}+2.\frac{b}{b+1}+2.\frac{c}{c+1}\ge5\sqrt[5]{\frac{ab^2c^2}{\left(a+1\right)\left(b+1\right)^2\left(c+1\right)^2}}\)

\(\Leftrightarrow\frac{1}{\left(c+1\right)^3}\ge125\sqrt[5]{\frac{a^3b^6c^6}{\left(a+1\right)^3\left(b+1\right)^6\left(c+1\right)^6}}\)

Nhân vế với vế:

\(\frac{1}{\left(a+1\right)\left(b+1\right)^2\left(c+1\right)^3}\ge5^6\sqrt[5]{\frac{a^5b^{10}c^{15}}{\left(a+1\right)^5\left(b+1\right)^{10}\left(c+1\right)^{15}}}=\frac{5^6ab^2c^3}{\left(a+1\right)\left(b+1\right)^2\left(c+1\right)^3}\)

\(\Leftrightarrow ab^2c^3\le\frac{1}{5^6}\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{5}\)

NV
16 tháng 11 2019

a/ Đề sai, đề đúng phải là \(p=\frac{a+b+c}{2}\)

b/ \(\Leftrightarrow\frac{2}{2+a^2b}+\frac{2}{2+b^2c}+\frac{2}{2+c^2a}\ge2\)

\(VT=1-\frac{a^2b}{1+1+a^2b}+1-\frac{b^2c}{1+1+b^2c}+1-\frac{c^2a}{1+1+c^2a}\)

\(VT\ge3-\left(\frac{a^2b}{3\sqrt[3]{a^2b}}+\frac{b^2c}{3\sqrt[3]{b^2c}}+\frac{c^2a}{3\sqrt[3]{c^2a}}\right)\)

\(VT\ge3-\frac{1}{9}\left(3\sqrt[3]{a^2.ab.ab}+3\sqrt[3]{b^2.bc.bc}+3\sqrt[3]{c^2.ca.ca}\right)\)

\(VT\ge3-\frac{1}{9}\left(a^2+2ab+b^2+2bc+c^2+2ca\right)\)

\(VT\ge3-\frac{1}{9}\left(a+b+c\right)^2=2\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)