Áp dụng BĐT Cauchy-Schwarz ta có:

\((ab+a+1)^2 \le (a+b+c) \left( a+ a^2b+ \frac 1c \right) = (a+b+c)(a+a^2b+ab)\)

\(\Rightarrow \dfrac{a}{(ab+a+1)^2} \ge \dfrac{a}{(a+b+c)(a+a^2b+ab)}= \dfrac{1}{(a+b+c)(1+ab+b)}\)

Thiết lập các BĐT tương tự rồi cộng theo vế ta có:

\(\sum \dfrac{a}{(ab+a+1)^2} \ge \dfrac{1}{a+b+c} \sum \dfrac{1}{ab+b+1}= \dfrac{1}{a+b+c}\)

c₂: Áp dụng BĐT bunyakovsky:

\(\left(a+b+c\right)\left[\dfrac{a}{\left(ab+a+1\right)^2}+\dfrac{b}{\left(bc+b+1\right)^2}+\dfrac{c}{\left(ac+c+1\right)^2}\right]\ge\left(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ca+c+1}\right)^2\)

Xét \(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}=\dfrac{a}{ab+a+1}+\dfrac{ab}{1+ab+a}+\dfrac{c}{c\left(a+1+ab\right)}\)

\(=\dfrac{ab+a+1}{ab+a+1}=1\)

do đó \(\left(a+b+c\right).VT\ge1\Leftrightarrow VT\ge\dfrac{1}{a+b+c}\)

dấu = xảy ra khi a=b=c=1

Ta có \(\sqrt{bc\left(1+a^2\right)}=\sqrt{bc+a^2bc}=\sqrt{bc+a\left(a+b+c\right)}\)

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

Đặt BT đề cho là P

\(\Leftrightarrow P=\sum\dfrac{a}{\sqrt{bc\left(1+a^2\right)}}=\sum\sqrt{\dfrac{a}{a+b}\cdot\dfrac{a}{a+c}}\\ \Leftrightarrow P\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{b+c}+\dfrac{b}{b+a}+\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)\\ \Leftrightarrow P\le\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{1}{2}\cdot3=\dfrac{3}{2}\)

Dấu \("="\Leftrightarrow a=b=c=\sqrt{3}\)

\(\left(a^2-bc\right)\left(b-abc\right)=\left(b^2-ca\right)\left(a-abc\right)\)

\(\Leftrightarrow a^2b+ab^2c^2-a^3bc-b^2c=b^2a+a^2bc^2-ca^2-ab^3c\)

\(\Leftrightarrow a^2b-ab^2-b^2c+ca^2=a^2bc^2-ab^3c+a^3bc-ab^2c^2\)

\(\Leftrightarrow\left(a-b\right)\left(ab+bc+ca\right)=abc\left(a-b\right)\left(a+b+c\right)\)

\(\Leftrightarrow ab+bc+ca=abc\left(a+b+c\right)\Leftrightarrow a+b+c=\dfrac{ab+bc+ca}{abc}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\left(đpcm\right)\)

Dễ dàng c/m : \(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}=1\)

Ta có : \(\dfrac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le\dfrac{1}{a+b+4}\le\dfrac{1}{4}\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}\right)\) 

Suy ra : \(\Sigma\dfrac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le2.\dfrac{1}{4}\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\right)=\dfrac{1}{2}.1=\dfrac{1}{2}\) 

" = " \(\Leftrightarrow a=b=c=1\)

\(3=ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}\Rightarrow abc\le1\)

\(\dfrac{1}{1+a^2\left(b+c\right)}=\dfrac{1}{1+a\left(ab+ac\right)}=\dfrac{1}{1+a\left(3-bc\right)}=\dfrac{1}{1+3a-abc}=\dfrac{1}{3a+\left(1-abc\right)}\le\dfrac{1}{3a}\)

Tương tự và cộng lại:

\(VT\le\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}=\dfrac{ab+bc+ca}{3abc}=\dfrac{3}{3abc}=\dfrac{1}{abc}\)

Bài 1:

Áp dụng BĐT AM-GM ta có:

$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 3\sqrt[3]{\frac{1}{(a+1)(b+1)(c+1)}}$

$\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\geq 3\sqrt[3]{\frac{abc}{(a+1)(b+1)(c+1)}}$

Cộng theo vế và thu gọn:

$\frac{a+1}{a+1}+\frac{b+1}{b+1}+\frac{c+1}{c+1}\geq \frac{3(1+\sqrt[3]{abc})}{\sqrt[3]{(a+1)(b+1)(c+1)}}$

$\Leftrightarrow 3\geq \frac{3(1+\sqrt[3]{abc})}{\sqrt[3]{(a+1)(b+1)(c+1)}}$

$\Rightarrow (a+1)(b+1)(c+1)\geq (1+\sqrt[3]{abc})^3$

Ta có đpcm.

Bài 2:

$a^3+a^3+a^3+a^3+b^3+c^3\geq 6\sqrt[6]{a^{12}b^3c^3}=6a^2\sqrt{bc}$

$b^3+b^3+b^3+b^3+a^3+c^3\geq 6b^2\sqrt{ac}$

$c^3+c^3+c^3+c^3+a^3+b^3\geq 6c^2\sqrt{ab}$

Cộng theo vế và rút gọn thu được:

$a^3+b^3+c^3\geq a^2\sqrt{bc}+b^2\sqrt{ac}+c^2\sqrt{ab}$ 

Ta có đpcm.

Dấu "=" xảy ra khi $a=b=c$

Bài 3:

Áp dụng BĐT Cauchy-Schwarz:

$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq \frac{(a+b+c)^2}{b+c+c+a+a+b}=\frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}$

Ta có đpcm

Dấu "=" xảy ra khi $a=b=c$

@Nguyễn Thanh Hằng đọc xong xóa đii nha