\(\dfrac{1}{2a-1}+\dfrac{1}{1}\ge\dfrac{4}{2a-1+1}=\dfrac{2}{a}\)

Tương tự: \(\dfrac{1}{2b-1}+1\ge\dfrac{2}{b}\) ; \(\dfrac{1}{2c-1}+1\ge\dfrac{2}{c}\)

Cộng vế:

\(VT\ge\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}=\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\ge\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

Lời giải:

Áp dụng BĐT Cauchy-Schwarz:

\(\text{VT}=\frac{\left ( \frac{a}{bc} \right )^2}{\frac{1}{c}}+\frac{\left ( \frac{b}{ca} \right )^2}{\frac{1}{a}}+\frac{\left ( \frac{c}{ab} \right )^2}{\frac{1}{b}}\geq \frac{\left ( \frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab} \right )^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\)

\(\Leftrightarrow \text{VT}\geq \frac{\left ( \frac{a^2+b^2+c^2}{abc} \right )^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\)

Theo hệ quả của BĐT AM-GM thì:

\(a^2+b^2+c^2\geq ab+bc+ac\)

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

Ta có đpcm.

Dấu bằng xảy ra khi \(a=b=c\)

Áp dụngk BĐt cô-si, ta có 

\(\frac{a^2}{b^2c}+\frac{b^2}{c^2a}+\frac{1}{a}\ge3.\frac{1}{c}\)

Tương tự , rồi cộng vào, ta có 

\(2A+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\Rightarrow A\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\left(ĐPCM\right)\)

^_^ 

a.

\(\sum\dfrac{ab}{a+c+b+c}\le\dfrac{1}{4}\sum\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)=\dfrac{a+b+c}{4}\)

2.

\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{a+b+2c+2b}\le\dfrac{ab}{9}\left(\dfrac{4}{a+b+2c}+\dfrac{1}{2b}\right)=4.\dfrac{ab}{a+b+2c}+\dfrac{a}{18}\)

Quay lại câu a

\(b,\dfrac{ab}{a+3b+2c}=\left(\dfrac{1}{9}ab\right)\cdot\dfrac{9}{\left(a+c\right)+\left(b+c\right)+2b}\le\left(\dfrac{1}{9}ab\right)\cdot\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{2b}\right)=\dfrac{1}{9}\cdot\left(\dfrac{ab}{a+b}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)

Cmtt: \(\dfrac{bc}{b+3c+2a}\le\dfrac{1}{9}\cdot\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+b}+\dfrac{b}{2}\right);\dfrac{ca}{c+3a+2b}\le\dfrac{1}{9}\cdot\left(\dfrac{ca}{b+c}+\dfrac{ca}{a+b}+\dfrac{c}{2}\right)\)

\(\Rightarrow VT\le\dfrac{1}{9}\left(\dfrac{bc+ca}{a+b}+\dfrac{ab+ac}{b+c}+\dfrac{ab+bc}{a+c}+\dfrac{a+b+c}{2}\right)\\ \le\dfrac{1}{9}\left(a+b+c+\dfrac{a+b+c}{2}\right)=\dfrac{1}{9}\cdot\dfrac{3}{2}\left(a+b+c\right)=\dfrac{a+b+c}{6}\)

Dấu $"="$ khi $a=b=c$

Do \(a+b+c=1\) nên Bất đẳng thức trên tương đương:
\(\dfrac{a}{1+a}+\dfrac{b}{1+b}+\dfrac{c}{1+c}\le\dfrac{3}{4}\)

\(\Leftrightarrow\left(1-\dfrac{1}{1+a}\right)+\left(1-\dfrac{1}{1+b}\right)+\left(1-\dfrac{1}{1+c}\right)\le\dfrac{3}{4}\)

\(\Leftrightarrow3-\left(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\right)\le\dfrac{3}{4}\)

Áp dụng BĐT cauchy-schwarz engel với a;b;c>0 ta có:

\(3-\left(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\right)\le3-\dfrac{\left(1+1+1\right)^2}{1+a+1+b+1+c}=3-\dfrac{9}{4}=\dfrac{3}{4}\)

Ta có:

\(\dfrac{a}{2a+b+c}+\dfrac{b}{a+2b+c}+\dfrac{c}{a+b+2c}=\dfrac{a}{\left(a+b\right)+\left(a+c\right)}+\dfrac{b}{\left(a+b\right)+\left(b+c\right)}+\dfrac{c}{\left(a+c\right)+\left(b+c\right)}=\dfrac{a}{4}.\dfrac{4}{\left(a+b\right)+\left(a+c\right)}+\dfrac{b}{4}.\dfrac{4}{\left(a+b\right)+\left(b+c\right)}+\dfrac{c}{4}.\dfrac{4}{\left(a+c\right)+\left(b+c\right)}=\dfrac{a}{4}.\dfrac{\left(1+1\right)^2}{\left(a+b\right)+\left(a+c\right)}+\dfrac{b}{4}.\dfrac{\left(1+1\right)^2}{\left(a+b\right)+\left(b+c\right)}+\dfrac{c}{4}.\dfrac{\left(1+1\right)^2}{\left(a+c\right)+\left(b+c\right)}\)Áp dụng BĐT Cauchy - Schwarz:

\(VT\le\dfrac{a}{4}.\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)+\dfrac{b}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)+\dfrac{c}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)=\dfrac{1}{4}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)=\dfrac{1}{4}.3=\dfrac{3}{4}\)\("="\Leftrightarrow a=b=c=\dfrac{1}{3}\)

Cho a,b,c >0, chứng minh rằng :\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\left(\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{... - Hoc24

Áp dụng bất đẳng thức \(\dfrac{9}{x+y+z}\le\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\) với x, y, z > 0 ta có:

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

\(4M=\dfrac{4}{\left(a+b\right)+\left(a+c\right)}+\dfrac{4}{\left(a+b\right)+\left(b+c\right)}+\dfrac{4}{\left(c+a\right)+\left(b+c\right)}\)

\(\le\dfrac{1}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{b+c}\)

\(=\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\)

=> 8M \(\le\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\)

\(\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a}=2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=8\)

=> \(M\le1\)

Dấu "=" xảy ra <=> a = b = c = 3/4 

\(\dfrac{1}{2a+b+c}=\dfrac{1}{a+a+b+c}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{16}\left(\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

Tương tự:

\(\dfrac{1}{a+2b+c}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{a+b+2c}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{c}\right)\)

Cộng vế:

\(M\le\dfrac{1}{16}\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\)

\(M_{max}=1\)  khi \(a=b=c=\dfrac{3}{4}\)

\(\dfrac{1}{a+3b}+\dfrac{1}{a+b+2c}\ge\dfrac{4}{2a+4b+2c}=\dfrac{2}{a+2b+c}\)

Tương tự: \(\dfrac{1}{b+3c}+\dfrac{1}{b+c+2a}\ge\dfrac{2}{a+b+2c}\)

\(\dfrac{1}{c+3a}+\dfrac{1}{a+c+2b}\ge\dfrac{2}{2a+b+c}\)

Cộng vế với vế và rút gọn:

\(\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{c+3a}\ge\dfrac{1}{a+2b+c}+\dfrac{1}{b+2c+a}+\dfrac{1}{c+2a+b}\)

Dấu "=" xảy ra khi \(a=b=c\)