BĐT cần chứng minh tương đương :

\(\sqrt{\dfrac{a^2+b^2}{2}}-\sqrt{ab}\ge\dfrac{a+b}{2}-\dfrac{2ab}{a+b}\)

\(\Leftrightarrow\dfrac{\dfrac{a^2+b^2}{2}-ab}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}\ge\dfrac{\left(a+b\right)^2-4ab}{2\left(a+b\right)}\)

\(\Leftrightarrow\dfrac{\dfrac{\left(a-b\right)^2}{2}}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}\ge\dfrac{\left(a-b\right)^2}{2\left(a+b\right)}\)

\(\Leftrightarrow\dfrac{\dfrac{\left(a-b\right)^2}{2}}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}-\dfrac{\left(a-b\right)^2}{2\left(a+b\right)}\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\left(\dfrac{\dfrac{1}{2}}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}-\dfrac{1}{2\left(a+b\right)}\right)\ge0\)

ta phải chứng minh;

\(\dfrac{\dfrac{1}{2}}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}-\dfrac{1}{2\left(a+b\right)}\ge0\)

\(\Leftrightarrow\)\(\dfrac{\dfrac{1}{2}}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}\ge\dfrac{1}{2\left(a+b\right)}\)

\(\Leftrightarrow a+b\ge\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\)\(\Leftrightarrow2a+2b-\sqrt{2\left(a^2+b^2\right)}-2\sqrt{ab}\ge0\)

\(\Leftrightarrow\left(a+b-\sqrt{2\left(a^2+b^2\right)}\right)+\left(a+b-2\sqrt{ab}\right)\ge0\)

\(\Leftrightarrow\dfrac{\left(a+b\right)^2-2\left(a^2+b^2\right)}{a+b+\sqrt{2\left(a^2+b^2\right)}}+\dfrac{\left(a+b\right)^2-4ab}{a+b+2\sqrt{ab}}\ge0\)

\(\Leftrightarrow\dfrac{-\left(a-b\right)^2}{a+b+\sqrt{2\left(a^2+b^2\right)}}+\dfrac{\left(a-b\right)^2}{a+b+2\sqrt{ab}}\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\left(\dfrac{1}{a+b+2\sqrt{ab}}-\dfrac{1}{a+b+\sqrt{2\left(a^2+b^2\right)}}\right)\ge0\)

ta phải chứng minh

\(\Leftrightarrow\dfrac{1}{a+b+2\sqrt{ab}}-\dfrac{1}{a+b+\sqrt{2\left(a^2+b^2\right)}}\ge0\)

\(\Leftrightarrow\dfrac{1}{a+b+2\sqrt{ab}}\ge\dfrac{1}{a+b+\sqrt{2\left(a^2+b^2\right)}}\)

\(\Leftrightarrow a+b+2\sqrt{ab}\le a+b+\sqrt{2\left(a^2+b^2\right)}\)

\(\Leftrightarrow2\sqrt{ab}\le\sqrt{2\left(a^2+b^2\right)}\Leftrightarrow\left(a-b\right)^2\ge0\)

BĐT cần chứng minh tương đương

\(\dfrac{3a^2+2ab+3b^2}{a+b}-2\left(a+b\right)\ge2\sqrt{2\left(a^2+b^2\right)}-2\left(a+b\right)\)

\(\Leftrightarrow\dfrac{a^2-2ab+b^2}{a+b}\ge\dfrac{8\left(a^2+b^2\right)-4\left(a+b\right)^2}{2\sqrt{2\left(a^2+b^2\right)}+2\left(a+b\right)}\)

\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{a+b}\ge\dfrac{2\left(a-b\right)^2}{\sqrt{2\left(a^2+b^2\right)}+a+b}\)

\(\Leftrightarrow\left(a-b\right)^2\left(\dfrac{1}{a+b}-\dfrac{2}{\sqrt{2\left(a^2+b^2\right)}+a+b}\right)\ge0\)

ta phải chứng minh

\(\dfrac{1}{a+b}-\dfrac{2}{\sqrt{2\left(a^2+b^2\right)}+a+b}\ge0\)

\(\Leftrightarrow\dfrac{1}{a+b}\ge\dfrac{2}{\sqrt{2\left(a^2+b^2\right)}+a+b}\)

\(\Leftrightarrow\sqrt{2\left(a^2+b^2\right)}+a+b\ge2\left(a+b\right)\Leftrightarrow\sqrt{2\left(a^2+b^2\right)}\ge a+b\)

\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\Leftrightarrow\left(a-b\right)^2\ge0\) (luôn đúng)

=> đpcm

Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)

\(\Rightarrow\left|x-1\right|+\left|x-3\right|=\left|x-1\right|+\left|3-x\right|\ge\left|2\right|=2\)

tương tự \(\left|x-2\right|+\left|x-4\right|\ge2\)

Cộng hai vế BĐT trên lại ta được

\(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+\left|x-4\right|\ge4\)

Vậy MinM=4 khi \(\left\{{}\begin{matrix}1\le x\le3\\2\le x\le4\end{matrix}\right.\)\(\Leftrightarrow2\le x\le3\)

Vi AB<BC va A,B,C thang hang suy ra Bnam giua A,C. Suy ra AC =AB+BC=2+3=5cm

Vay khoang.cach tu A den C la5cm.           De ot.  Dung quen tick cho t

nghiệm hay là nghiệm nguyên ?

có nghĩa là có \(n\) số hạng \(\dfrac{1}{n}\) đó bạn

ví dụ : có 2003 số hạng \(\dfrac{1}{2003}\) thì sẽ được viết như đề chỉ thay n=2003 thôi

Áp dụng BĐT B.C.S ta có :

\(\dfrac{1}{2ab^2+1}+\dfrac{1}{2bc^2+1}+\dfrac{1}{2ca^2+1}\ge\dfrac{9}{2ab^2+2bc^2+2ca^2+3}\)

Ta phải chứng minh \(\dfrac{9}{2ab^2+2bc^2+2ca^2+3}\ge1\)

\(\Leftrightarrow2ab^2+2bc^2+2ac^2+3\le9\) do a,b,c dương nên chia cả hai vế cho abc ta được: \(2\left(a+b+c\right)+\dfrac{3}{abc}\le\dfrac{9}{abc}\)

\(\Leftrightarrow6\le\dfrac{6}{abc}\Leftrightarrow abc\le1\) Bất đẳng thức cuối luôn đúng thật vậy:

áp dụng BĐT AM - GM :

\(\Rightarrow a+b+c\ge3\sqrt[3]{abc}\Leftrightarrow abc\le1\)

\(\Rightarrowđpcm\)

c)

ta có : \(\dfrac{1}{1+x}\ge1-\dfrac{1}{1+y}+1-\dfrac{1}{1+z}=\dfrac{y}{y+1}+\dfrac{z}{z+1}\)

Áp dụng bất đẳng thức AM - GM :

\(\Rightarrow\dfrac{1}{1+x}\ge\dfrac{y}{y+1}+\dfrac{z}{z+1}\ge2\sqrt{\dfrac{yz}{\left(y+1\right)\left(z+1\right)}}\)

tương tự ta cũng có

\(\dfrac{1}{1+y}\ge2\sqrt{\dfrac{xz}{\left(x+1\right)\left(z+1\right)}}\) ; \(\dfrac{1}{1+z}\ge2\sqrt{\dfrac{xy}{\left(x+1\right)\left(y+1\right)}}\)

nhân ba vế bất đẳng thức lại với nhau ta được

\(\dfrac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge8\sqrt{\dfrac{x^2y^2z^2}{\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2}}\)

\(\Rightarrow\dfrac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge\dfrac{8xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)

\(\Rightarrow xyz\le\dfrac{1}{8}\) \(\Rightarrow Max\left(xyz\right)=\dfrac{1}{8}\) khi \(x=y=z=\dfrac{1}{2}\)