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 $a=b=c$

\(vì:a,b,c>0\Rightarrow\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}>0\)

\(Cosi:\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{2}{\sqrt{ab}}\ge\dfrac{2}{\dfrac{a+b}{2}}=\dfrac{4}{a+b}\)

\(\dfrac{4}{2a+b+c}\le\dfrac{1}{4}\left(\dfrac{4}{a+b}+\dfrac{4}{a+c}\right)\le\dfrac{1}{16}\left(\dfrac{8}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=\dfrac{1}{2a}+\dfrac{1}{4b}+\dfrac{1}{4c}.tươngtự:\dfrac{4}{a+b+2c}\le\dfrac{1}{4a}+\dfrac{1}{4b}+\dfrac{1}{2c};\dfrac{4}{a+2b+c}\le\dfrac{1}{4a}+\dfrac{1}{2b}+\dfrac{1}{2c}.\text{cộng vế theo vế ta được:}\dfrac{4}{a+2b+c}+\dfrac{4}{2a+b+c}+\dfrac{4}{a+b+2c}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\left(\text{đpcm}\right)\)

Áp dụng BĐT \(\dfrac{1}{x+y+z+t}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{t}\right)\) với các số dương

Ta có: \(\dfrac{4}{a+a+b+c}\le\dfrac{4}{16}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

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

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

Cộng vế với vế:

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

Dấu "=" xảy ra khi \(a=b=c\)

* Ta cm bđt : \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\forall ab\)

+ \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)

\(\Leftrightarrow\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\)

\(\Leftrightarrow\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow\left(a+b\right)^2-4ab\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\ge0\)

Vì bđt thức cuối luôn đúng mà các phép biến đổi trên là tương đương nên ta có đpcm

Dấu "=" \(\Leftrightarrow x=y\)

+ Áp dụng bđt \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)

Dấu "=" \(\Leftrightarrow x=y\) ta có :

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) Dấu "=" xảy ra \(\Leftrightarrow a=b\)

\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\) Dấu "=" xảy ra \(\Leftrightarrow b=c\)

\(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{a+c}\) Dấu "=" xảy ra \(\Leftrightarrow a=c\)

Do đó : \(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge4\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge2\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\)

+ Áp dụng bđt trên một lần nữa ta có :

\(\dfrac{1}{a+b}+\dfrac{1}{b+c}\ge\dfrac{4}{a+2b+c}\) Dấu "=" xảy ra \(\Leftrightarrow a=c\)

\(\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge\dfrac{1}{a+b+2c}\) Dấu "=" xảy ra \(\Leftrightarrow a=b\)

\(\dfrac{1}{a+b}+\dfrac{1}{c+a}\ge\dfrac{4}{2a+b+c}\) Dấu "=" xảy ra \(\Leftrightarrow b=c\)

Do đó : \(2\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge\dfrac{4}{2a+b+c}\)

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

=> đpcm

Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

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

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 \(a=b=c\)

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

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 \(a=b=c\)

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 

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

^_^ 

Bài 2:

\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)

Trước hết ta chứng minh \(\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)

Áp dụng BĐT AM-GM ta có:

\(\sqrt{a\left(b+c\right)}\le\dfrac{a+b+c}{2}\)\(\Rightarrow1\ge\dfrac{2\sqrt{a\left(b+c\right)}}{a+b+c}\)

\(\Rightarrow\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\). Ta lại có:

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

Thiết lập các BĐT tương tự:

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

Cộng theo vế 3 BĐT trên ta có:

\(VT\ge\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}\ge2\)

Dấu "=" không xảy ra nên ta có ĐPCM

Lưu ý: lần sau đăng từng bài 1 thôi nhé !

1) Áp dụng liên tiếp bđt \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) với a;b là 2 số dương ta có:

\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{\dfrac{1}{a+b}+\dfrac{1}{a+c}}{4}\)\(\le\dfrac{\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}}{16}\)

TT: \(\dfrac{1}{a+2b+c}\le\dfrac{\dfrac{2}{b}+\dfrac{1}{a}+\dfrac{1}{c}}{16}\)

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

Cộng vế với vế ta được:

\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)

Ai lm dc bai 3 chua

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:

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