Áp dụng bất đẳng thức cho hai số dương

\(\dfrac{1}{\left(a+b\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

Xét \(c+1=c+a+b+c\)

\(\dfrac{ab}{c+1}\le\dfrac{ab}{4\left[\dfrac{1}{a+c}+\dfrac{1}{b+c}\right]}\)

Tương tự:

\(\dfrac{bc}{a+1}\le\dfrac{bc}{4\left[\dfrac{1}{a+c}+\dfrac{1}{b+a}\right]}\)

\(\dfrac{ca}{b+1}\le\dfrac{ac}{4\left[\dfrac{1}{a+b}+\dfrac{1}{c+b}\right]}\)

Cộng lại :
\(\dfrac{ab}{c+1}+\dfrac{bc}{a+1}+\dfrac{ca}{b+1}\le\dfrac{1}{4}\left[\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{bc}{a+c}+\dfrac{bc}{a+b}+\dfrac{ac}{a+b}+\dfrac{ac}{b+c}\right]\)

Rút gọn mẫu số
\(\Rightarrow\dfrac{ab}{c+1}+\dfrac{bc}{a+1}+\dfrac{ca}{b+1}\le\dfrac{1}{4}\left(a+b+c\right)=\dfrac{1}{4}\)

Ta có: \(0\le a\le b\le1.\)

\(\Rightarrow\left\{{}\begin{matrix}a-1\le0\\b-1\le0\end{matrix}\right.\)

\(\Rightarrow\left(a-1\right).\left(b-1\right)\ge0\)

\(\Rightarrow ab-a-b+1\ge0.\)

\(\Rightarrow ab+1\ge0+a+b\)

\(\Rightarrow ab+1\ge a+b\)

\(\Rightarrow\frac{1}{ab+1}\le\frac{1}{a+b}.\)

\(\Rightarrow\frac{c}{ab+1}\le\frac{c}{a+b}\left(c\ge0\right).\)

Mà \(\frac{c}{a+b}\le\frac{2c}{a+b+c}\left(c\ge0\right)\)

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

Chứng minh tương tự ta cũng có:

\(\frac{b}{ac+1}\le\frac{2b}{a+b+c}\left(2\right);\frac{a}{bc+1}\le\frac{2a}{a+b+c}\left(3\right).\)

Cộng theo vế \(\left(1\right);\left(2\right)và\left(3\right)\) ta được:

\(\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\le\frac{2a}{a+b+c}+\frac{2b}{a+b+c}+\frac{2c}{a+b+c}\)

\(\Rightarrow\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\le\frac{2a+2b+2c}{a+b+c}\)

\(\Rightarrow\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\le\frac{2.\left(a+b+c\right)}{a+b+c}\)

\(\Rightarrow\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\le2\left(đpcm\right).\)

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Cảm ơn bạn nha!

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