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

\(\left(1+a\right)\left(1+b\right)+\left(1+b\right)\left(1+c\right)+\left(1+c\right)\left(1+a\right)\ge5\)

\(\Leftrightarrow1+a+b+ab+1+b+c+bc+1+c+a+ca\ge5\)

\(\Leftrightarrow3+2\left(a+b+c\right)+ab+bc+ca\ge5\)

\(\Leftrightarrow5+ab+bc+ca\ge5\)(luôn đúng)

Dấu "=" xảy ra khi a=b=0;c=1 và các hoán vị

Ta có :

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

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

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

Áp dụng BĐT Cô-si,ta có :

\(\frac{a+b}{a}+\frac{a+c}{a}\ge2\sqrt{\frac{\left(a+b\right)\left(a+c\right)}{a^2}}\)

\(\frac{a+b}{b}+\frac{b+c}{b}\ge2\sqrt{\frac{\left(a+b\right)\left(b+c\right)}{b^2}}\)

\(\frac{a+c}{c}+\frac{b+c}{c}\ge2\sqrt{\frac{\left(a+c\right)\left(b+c\right)}{c^2}}\)

\(\Rightarrow\left(\frac{a+b}{a}+\frac{a+c}{a}\right)\left(\frac{a+b}{b}+\frac{b+c}{b}\right)\left(\frac{a+c}{c}+\frac{b+c}{c}\right)\ge8\sqrt{\frac{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}{a^2b^2c^2}}\)

\(\ge8\sqrt{\frac{\left[8\sqrt{a^2b^2c^2}\right]^2}{a^2b^2c^2}}=8\sqrt{64}=64\)

Dấu "=" xảy ra khi a = b = c = \(\frac{1}{3}\)

\(\left(\dfrac{1}{a}-1\right)\left(\dfrac{1}{b}-1\right)\left(\dfrac{1}{c}-1\right)=\left(\dfrac{1-a}{a}\right)\left(\dfrac{1-b}{b}\right)\left(\dfrac{1-c}{c}\right)\)

\(=\left(\dfrac{b+c}{a}\right)\left(\dfrac{a+c}{b}\right)\left(\dfrac{a+b}{c}\right)\ge\dfrac{2\sqrt{bc}}{a}.\dfrac{2\sqrt{ac}}{b}.\dfrac{2\sqrt{ab}}{c}=8\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)