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 $a=b=c$

Dấu >= hay <= vậy bạn? Bạn xem lại đề.

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

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 \(a=b=c\)

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

Tương tự:

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

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

Cộng vế:

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

Dấu "=" xảy ra khi \(a=b=c\)

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

Giả sử c là số ở giửa a và b. khi đó \(\left(b-c\right)\left(c-a\right)\ge0\)

Ta chứng minh :

\(VT\le c\left(\dfrac{b^2}{2b^2+a^2+c^2}+\dfrac{a^2}{2a^2+b^2+c^2}\right)+\dfrac{abc}{a^2+b^2+2c^2}\)(*)

\(\Leftrightarrow\dfrac{\left(c-a\right)\left(b-c\right)\left(b^2+c^2-bc+a^2\right)}{\left(a^2+c^2+2b^2\right)\left(b^2+a^2+2c^2\right)}\ge0\) (Đúng)

Áp dụng BĐT AM-GM:

\(VT\le\dfrac{c}{4}\left(\dfrac{b^2}{a^2+b^2}+\dfrac{b^2}{b^2+c^2}+\dfrac{a^2}{a^2+b^2}+\dfrac{a^2}{a^2+c^2}\right)+\dfrac{abc}{2ac+2bc}\)

\(\le\dfrac{c}{4}\left(1+\dfrac{b^2}{2bc}+\dfrac{a^2}{2ac}\right)+\dfrac{\dfrac{\left(a+b\right)^2}{4}}{2\left(a+b\right)}=\dfrac{c}{4}+\dfrac{a+b}{8}+\dfrac{a+b}{8}\)

\(=\dfrac{a+b+c}{4}\)( \(ĐpcM\))

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

Ta có:

\(a^2+b^2\ge2ab\)

\(b^2+1\ge2ab\)

\(\Rightarrow a^2+2ab^2+3\ge2\left(ab+b+1\right)\)

\(\Rightarrow\dfrac{1}{a^2+2b^2+3}< \dfrac{1}{2.\left(ab+b+1\right)}\)

Tương tự:

\(\dfrac{1}{a^2+2b^2+3}+\dfrac{1}{b^2+2c^2+3}+\dfrac{1}{c^2+2a^2+3}\le\dfrac{1}{2}.\left(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ca+a+1}\right)\)

Mặt khác:

\(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ca+a+1}=\dfrac{1}{ab+b+1}+\dfrac{ab}{ab^2c+abc+ab}+\dfrac{b}{bca+ab+b}=1\)

\(\Rightarrow\dfrac{1}{a^2+2b^2+3}+\dfrac{1}{b^2+2c^2+3}+\dfrac{1}{c^2+2a^2+3}\le\dfrac{1}{2}\)

\(\Leftrightarrow a=b=c=1\)

\(\Rightarrow\) Đpcm.

Áp dụng BĐT AM - GM, ta có:

\(a^2+2b^2+3\)

\(=\left(a^2+b^2\right)+\left(b^2+1\right)+2\)

\(\ge2ab+2b+2\)

Tương tự, ta có: \(b^2+2c^2+3\ge2bc+2c+2\) và \(c^2+2a^2+3\ge2ac+2a+2\)

\(VT=\dfrac{1}{a^2+2b^2+3}+\dfrac{1}{b^2+2c^2+3}+\dfrac{1}{c^2+2a^2+3}\)

\(\le\dfrac{1}{2ab+2b+2}+\dfrac{1}{2bc+2c+2}+\dfrac{1}{2ac+2a+2}\)

\(=\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ac+a+1}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{abc}{bc+c+abc}+\dfrac{abc}{ac+a^2bc+abc}\right)\) (Thay abc = 1)

\(=\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{ab}{b+1+ab}+\dfrac{b}{1+ab+b}\right)\)

\(=\dfrac{1}{2}\times\dfrac{1+ab+b}{ab+b+1}\)

\(=\dfrac{1}{2}=VP\left(\text{đ}pcm\right)\)

Dấu "=" xảy ra khi a = b = c = 1

Ta có: \(P\le\dfrac{a}{2a+2b+2}+\dfrac{b}{2b+2c+2}+\dfrac{c}{2c+2a+2}\)

Nên ta chỉ cần chứng minh:

\(\dfrac{a}{a+b+1}+\dfrac{b}{b+c+1}+\dfrac{c}{c+a+1}\le1\)

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

\(\Leftrightarrow\dfrac{b+1}{a+b+1}+\dfrac{c+1}{b+c+1}+\dfrac{a+1}{c+a+1}\ge2\)

Thật vậy, ta có:

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

\(VT\ge\dfrac{\left(a+b+c+3\right)^2}{ab+bc+ca+3\left(a+b+c\right)+6}=\dfrac{2\left(ab+bc+ca\right)+6\left(a+b+c\right)+12}{ab+bc+ca+3\left(a+b+c\right)+6}=2\)

Dấu "=" xảy ra khi \(a=b=c=1\)