xét hiệu \(\frac{1}{a}+\frac{1}{b}-\frac{4}{a+b}=\frac{a+b}{ab}-\frac{4}{a+b}\)

\(=\frac{\left(a+b\right)^2}{ab\left(a+b\right)}-\frac{4ab}{ab\left(a+b\right)}\)

\(=\frac{a^2+2ab+b^2-4ab}{ab\left(a+b\right)}\)

\(=\frac{\left(a-b\right)^2}{ab\left(a+b\right)}\)

vì (a-b)²>=0

mà a,b>0 nên ab>0;a+b>0

\(\Rightarrow\frac{\left(a-b\right)^2}{ab\left(a+b\right)}\ge0\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}-\frac{4}{ab}\ge0\)

hay \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{ab}\left(dpcm\right)\)

1/a+1/b>=2/căn ab

a+b>=2căn ab

=>(1/a+1/b)(a+b)>=4

Áp dụng BĐT Shur  ta có: \(\frac{1}{a}+\frac{1}{b}\ge\)\(\frac{\left(1+1\right)^2}{a+b}\)=\(\frac{4}{a+b}\)

Dấu = khi a=b

dấu "='' xảy ra khi a=b

Bài 1: Theo đề bài: \(VT=\left(a-1\right)+\frac{1}{\left(a-1\right)}+1\ge2\sqrt{\left(a-1\right).\frac{1}{a-1}}+1=2+1=3^{\left(đpcm\right)}\)

Đẳng thức xảy ra khi \(\left(a-1\right)=\frac{1}{a-1}\Leftrightarrow a=2\)

Bài 2: \(BĐT\Leftrightarrow\left(a^2+2\right)^2\ge4\left(a^2+1\right)\)

\(\Leftrightarrow a^4+4a^2+4\ge4a^2+4\)

\(\Leftrightarrow a^4\ge0\) (đúng). Đẳng thức xảy ra khi a = 0

Bài 3: Hình như sai đề thì phải ạ. Nếu a = 1,5 ; b = 1 thì \(\frac{19}{10}=1,9< 3\)

BĐT svac

\(\frac{1}{a}+\frac{1}{b}\ge\frac{\left(1+1\right)^2}{a+b}=\frac{4}{a+b}\forall a,b>0\)

Bài 1:

a) Áp dụng BĐT Cô-si:

\(VT=a-1+\frac{1}{a-1}+1\ge2\sqrt{\frac{a-1}{a-1}}+1=2+1=3\)

Dấu "=" xảy ra \(\Leftrightarrow a=2\).

b) BĐT \(\Leftrightarrow a^2+2\ge2\sqrt{a^2+1}\)

\(\Leftrightarrow a^2+1-2\sqrt{a^2+1}+1\ge0\)

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

Dấu "=" xảy ra \(\Leftrightarrow a=0\).

Bài 2: tương tự 1b.

Bài 3:

Do \(a,b,c\) dương nên ta có các BĐT:

\(\frac{a}{a+b+c}< \frac{a}{a+b}< \frac{a+c}{a+b+c}\)

Tương tự: \(\frac{b}{a+b+c}< \frac{b}{b+c}< \frac{b+a}{a+b+c};\frac{c}{a+b+c}< \frac{c}{c+a}< \frac{c+b}{a+b+c}\)

Cộng theo vế 3 BĐT:

\(\frac{a+b+c}{a+b+c}< \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< \frac{2\left(a+b+c\right)}{a+b+c}\)

\(\Leftrightarrow1< \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< 2\)( đpcm )

Ta có:

\(\dfrac{3}{a}+\dfrac{3}{b}\ge\dfrac{12}{a+b}\) (1)

\(\Leftrightarrow\dfrac{3a\left(a+b\right)+3b\left(a+b\right)-12ab}{ab\left(a+b\right)}\ge0\)

\(\Leftrightarrow\dfrac{3a^2+3ab+3ab+3b^2-12ab}{ab\left(a+b\right)}\ge0\)

\(\Leftrightarrow\dfrac{3a^2+3b^2-6ab}{ab\left(a+b\right)}\ge0\)

\(\Leftrightarrow\dfrac{3\left(a-b\right)^2}{ab\left(a+b\right)}\ge0\) ( luôn đúng)

Tương tự ta có:

\(\dfrac{2}{b}+\dfrac{2}{c}\ge\dfrac{8}{b+c}\) (2)

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

Cộng vế (1) (2)(3) ta được:

\(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{2}{b}+\dfrac{2}{c}+\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{12}{a+b}+\dfrac{8}{b+c}+\dfrac{4}{c+a}\)

\(\Leftrightarrow\dfrac{4}{a}+\dfrac{5}{b}+\dfrac{3}{c}\ge4\left(\dfrac{3}{a+b}+\dfrac{2}{b+c}+\dfrac{1}{c+a}\right)\)