Áp dụng giả thiết \(ab=1\) và bất đẳng thức Cauchy ta có:

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

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}ab=1\\a-b=\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{\sqrt{6}+\sqrt{2}}{2}\\b=\dfrac{\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\)

Lời giải:

Áp dụng BĐT Cô-si ta có:

$\frac{a^2+b^2}{a-b}=\frac{(a-b)^2+2ab}{a-b}=\frac{(a-b)^2+2}{a-b}=(a-b)+\frac{2}{a-b}\geq 2\sqrt{(a-b).\frac{2}{a-b}}=2\sqrt{2}$

Ta có đpcm.

\(\dfrac{a}{bc}+\dfrac{b}{ac}>=2\cdot\sqrt{\dfrac{a}{bc}\cdot\dfrac{b}{ac}}=\dfrac{2}{cc}\)

\(\dfrac{b}{ca}+\dfrac{c}{ab}>=2\cdot\sqrt{\dfrac{bc}{ca\cdot ab}}=\dfrac{2}{a}\)

\(\dfrac{c}{ab}+\dfrac{a}{bc}>=2\cdot\sqrt{\dfrac{a\cdot c}{a\cdot b\cdot c\cdot b}}=\dfrac{2}{b}\)

=>a/bc+b/ac+c/ab>=2(1/a+1/b+1/c)

Chỗ kia là có thêm dấu + nữa nha

*Cách khác

Khá căn bản thôi áp dụng BĐt cosi với 2 số dương

`=>a+(b+c)>=2sqrt{a(b+c)}`

`=>a/(2sqrt{a(b+c)})>=a/(a+b+c)`

`<=>sqrt{a/(b+c)}>=(2a)/(a+b+c)`

CMTT:

`sqrt{b/(c+a)}>=(2b)/(a+b+c)`

`sqrt{c/(a+b)}>=(2c)/(a+b+c)`

`=>sqrt{a/(b+c)}+sqrt{b/(c+a)}+sqrt{c/(a+b)}>=2`

Dấu "=" `<=>a=b=c=0` vô lý vì `a,b,c>0`

a/ Xét hiệu: \(a+b\ge2\sqrt{ab}\)

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

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)(luôn đúng) (đpcm)

''='' xảy ra khi a = b

b/ Sửa đề chút nhé: CMR:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{1}{\sqrt{ab}}+\dfrac{1}{\sqrt{bc}}+\dfrac{1}{\sqrt{ac}}\)

Áp dụng bđt AM-GM có:

\(\dfrac{1}{a}+\dfrac{1}{b}\ge2\sqrt{\dfrac{1}{a}\cdot\dfrac{1}{b}}=2\sqrt{\dfrac{1}{ab}}=\dfrac{2}{\sqrt{ab}}\);

Tương tự ta có:

\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{\sqrt{bc}}\); \(\dfrac{1}{a}+\dfrac{1}{c}\ge\dfrac{2}{\sqrt{ac}}\)

Cộng 2 vế ba bđt trên ta được:

\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge2\left(\dfrac{1}{\sqrt{ab}}+\dfrac{1}{\sqrt{bc}}+\dfrac{1}{\sqrt{ac}}\right)\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{1}{\sqrt{ab}}+\dfrac{1}{\sqrt{bc}}+\dfrac{1}{\sqrt{ac}}\left(đpcm\right)\)

''='' xảy ra khi a = b = c

a) Áp dụng BĐT Cauchy cho 2 số không âm , ta có:

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

\(\Rightarrow a+b\ge2\sqrt{ab}\)

b) Xét hiệu:

\(\dfrac{a}{b}+\dfrac{b}{a}-2=\dfrac{a^2+b^2-2ab}{ab}=\dfrac{\left(a-b\right)^2}{ab}\ge0\) ( luôn đúng)

=> \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\)

a) a + b ≥ 2\(\sqrt{ab}\) ( a > 0 ; b > 0 )

⇔ a - 2\(\sqrt{ab}\) + b ≥ 0

⇔ \(\left(\sqrt{a}-\sqrt{b}\right)^2\) ≥ 0 ( luôn đúng )

b) Áp dụng BĐT Cô-si :

x² + y² ≥ 2xy ( x > 0 ; y > 0)

⇒ a² + b² ≥ 2ab ( a > 0 ; b > 0)

⇔ \(\dfrac{a^2+b^2}{ab}\) ≥ 2

⇔\(\dfrac{a}{b}+\dfrac{b}{a}\) ≥ 2

Câu a :

Ta có :

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

\(\Rightarrow a-2\sqrt{ab}+b\ge0\)

\(\Rightarrow a+b\ge2\sqrt{ab}\)

Dấu \("="\) xảy ra khi \(a=b\)

Câu b :

\(\dfrac{a}{b}+\dfrac{b}{a}=\dfrac{a^2+b^2}{ab}\)

Ta có :

\(\dfrac{a^2+b^2}{ab}\ge2\Leftrightarrow a^2+b^2\ge2ab\Leftrightarrow\left(a-b\right)^2\ge0\) ( đúng )

Ta có: \(\dfrac{a+b}{2}\ge\sqrt{ab}\)

\(\Leftrightarrow a+b\ge2\sqrt{ab}\)

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

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)(luôn đúng)

*Chứng minh bất đẳng thức

Ta có: \(\forall a,b\ge0\) thì \(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)

\(\Leftrightarrow a+b-2\sqrt{ab}\ge0\) \(\Leftrightarrow a+b\ge2\sqrt{ab}\) \(\Leftrightarrow\dfrac{a+b}{2}\ge\sqrt{ab}\)  (đpcm)

Ta có: \(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\forall a,b>0\)

\(\Leftrightarrow a-2\sqrt{ab}+b\ge0\forall a,b>0\)

\(\Leftrightarrow a+b\ge2\sqrt{ab}\forall a,b>0\)

\(\Leftrightarrow\dfrac{a+b}{2}\ge\sqrt{ab}\forall a,b>0\)(đpcm)

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

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

\(\ge\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{4c}{a+b+2c}=\dfrac{\left(a-b\right)^2\left(a+b+c\right)}{\left(b+c\right)\left(c+a\right)\left(a+b+2c\right)}\ge0\)

(đúng hiển nhiên)

Đẳng thức xảy ra khi $a=b=c.$