Bài 5.

1. Chứng minh

$\dfrac{2}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}$

Ta có:

$\dfrac{2}{a}+\dfrac{1}{b}-\dfrac{4}{a+b}$

$=\dfrac{2b(a+b)+a(a+b)-4ab}{ab(a+b)}$

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

$=\dfrac{(a-b)^2+b^2}{ab(a+b)}\ge0$

Vậy: $\boxed{\dfrac{2}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}}$

2. Chứng minh

$\dfrac1a+\dfrac1b+\dfrac1c\ge\dfrac{a}{a+b+c}$

Vì $a,b,c>0$ nên:

$\dfrac1a+\dfrac1b+\dfrac1c>\dfrac1a$

Mà: $\dfrac1a>\dfrac{a}{a+b+c}$

Suy ra: $\boxed{\dfrac1a+\dfrac1b+\dfrac1c\ge\dfrac{a}{a+b+c}}$

1.

$a^3+b^4-ab(a+b)$

$=a^3+b^4-a^2b-ab^2$

$=a^2(a-b)+b^2(b-a)$

$=(a-b)(a^2-b^2)$

$=(a-b)^2(a+b)\ge0$

Suy ra: $\boxed{a^3+b^4\ge ab(a+b)}$

2.

$a^4+b^4-ab(a^2+b^2)$

$=a^4+b^4-a^3b-ab^3$

$=a^3(a-b)+b^3(b-a)$

$=(a-b)(a^3-b^3)$

$=(a-b)^2(a^2+ab+b^2)\ge0$

Vậy: $\boxed{a^4+b^4\ge ab(a^2+b^2)}$

3.

$a^5+b^5-ab(a^3+b^3)$

$=a^5+b^5-a^4b-ab^4$

$=a^4(a-b)+b^4(b-a)$

$=(a-b)(a^4-b^4)$

$=(a-b)^2(a+b)(a^2+b^2)\ge0$

Vậy: $\boxed{a^5+b^5\ge ab(a^3+b^3)}$

Áp dụng BĐT cô si ta có:

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

\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)}\le\frac{a^2+1+b^2+1}{2}=\frac{a^2+b^2+2}{2}\)

\(\Rightarrow\frac{1}{a^2+1}+\frac{1}{b^2+1}\ge\frac{4}{a^2+b^2+2}\left(1\right)\)

Ta thấy: \(\frac{4}{a^2+b^2+2}=1-\frac{a^2+b^2-2}{a^2+b^2+2}\left(2\right)\)

Ta có: \(a^2+b^2+2\ge2ab+2=2\left(ab+1\right)\)

\(\Rightarrow\frac{1}{a^2+b^2+2}\le\frac{1}{2\left(ab+1\right)}\)

\(\Rightarrow\frac{-1}{a^2+b^2+2}\ge\frac{-1}{2\left(ab+1\right)}\)

Mà \(a^2+b^2-2\ge2\left(ab-1\right)\)

\(\Rightarrow1-\frac{a^2+b^2-2}{a^2+b^2+2}\ge1-\frac{2\left(ab-1\right)}{2\left(ab+1\right)}\)

\(=1-\frac{ab-1}{ab+1}=\frac{ab+1-ab+1}{ab+1}=\frac{2}{ab+1}\left(3\right)\)

Từ \(\left(1\right);\left(2\right)\) và \(\left(3\right)\) suy ra:

\(\frac{1}{a^2+1}+\frac{1}{b^2+1}\ge\frac{2}{ab+1}\) (Đpcm)

ta có : \(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}\ge\dfrac{2}{1+ab}\) \(\Leftrightarrow\left(\dfrac{1}{1+a^2}-\dfrac{1}{1+b^2}\right)+\left(\dfrac{1}{1+b^2}-\dfrac{1}{1+ab}\right)\ge0\)

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

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

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

bất đẳng thức này đúng vì ab\(\ge\) 1