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: $\boxed{a^5+b^5\ge ab(a^3+b^3)}$

b)

Đề: Cho a, b, c > 0 và abc = ab + bc + ca. Chứng minh rằng: \(\frac{1}{a+2b+3c}+\frac{1}{2a+3b+c}+\frac{1}{3a+b+2c}\le\frac{3}{16}\)

~ ~ ~ ~ ~

\(abc=ab+bc+ca\)

\(\Leftrightarrow1=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

Áp dụng BĐT \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\), ta có:

\(\frac{1}{a+2b+3c}+\frac{1}{2a+3b+c}+\frac{1}{3a+b+2c}\)

\(\le\frac{1}{4}\left(\frac{1}{a+c}+\frac{1}{2\left(b+c\right)}+\frac{1}{2\left(a+b\right)}+\frac{1}{b+c}+\frac{1}{2\left(a+c\right)}+\frac{1}{a+b}\right)\)

\(=\frac{1}{4}\left[\frac{3}{2\left(a+c\right)}+\frac{3}{2\left(b+c\right)}+\frac{3}{2\left(a+b\right)}\right]\)

\(=\frac{3}{8}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{a+b}\right)\)

\(\le\frac{3}{32}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(=\frac{3}{16}\) (đpcm)

Dấu "=" xảy ra khi a = b = c 

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