a)Áp dụng Bđt Cô si ta có:

\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge\frac{3}{\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)

\(\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\ge\frac{3\sqrt[3]{abc}}{\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)

Cộng theo vế 2 bđt trên ta có:

\(3\ge\frac{3\left(\sqrt[3]{abc}+1\right)}{\sqrt[3]{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)\(\Rightarrow\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+\sqrt[3]{abc}\right)^3\)

Dấu = khi a=b=c

b)Áp dụng Bđt Cô-si ta có:

\(\frac{bc}{a}+\frac{ca}{b}\ge2\sqrt{\frac{bc^2a}{ab}}=2c\)

\(\frac{ca}{b}+\frac{ab}{c}\ge2\sqrt{\frac{ca^2b}{bc}}=2a\)

\(\frac{bc}{a}+\frac{ab}{c}\ge2\sqrt{\frac{b^2ac}{ac}}=2b\)

Cộng theo vế 3 bđt trên ta có:

\(2\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)\ge2\left(a+b+c\right)\)

\(\Rightarrow\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\ge a+b+c\)

Đấu = khí a=b=c

bn sử đấu = khí là dấu = khi nhé

Lời giải:

Áp dụng BĐT Schur bậc 3 ta có:

$abc\geq (a+b-c)(b+c-a)(c+a-b)=(3-2c)(3-2a)(3-2b)$

$\Leftrightarrow abc\geq 12(ab+bc+ac)-18(a+b+c)+27-8abc$

$\Leftrightarrow 9abc\geq 12(ab+bc+ac)-27$

$\Leftrightarrow abc\geq \frac{4}{3}(ab+bc+ac)-3$

$\Rightarrow 2abc\geq \frac{8}{3}(ab+bc+ac)-6(*)$

Mặt khác:

$\frac{8}{3}(ab+bc+ac)-6-[3(ab+bc+ac)-7]=1-\frac{ab+bc+ac}{3}$

$=\frac{(a+b+c)^2}{9}-\frac{ab+bc+ac}{3}=\frac{(a-b)^2+(b-c)^2+(c-a)^2}{9}\geq 0$

$\Rightarrow \frac{8}{3}(ab+bc+ac)-6\geq 3(ab+bc+ac)-7(**)$

Từ $(*); (**)\Rightarrow 2abc\geq 3(ab+bc+ac)-7$

$\Rightarrow 3(ab+bc+ac)-2abc\leq 7$

Dấu "=" xảy ra khi $a=b=c=1$ (vô lý vì $c>\frac{3}{2}$)

Do đó dấu "=" không xảy ra nên $3(ab+bc+ac)-2abc< 7$ (đpcm)

\(a^2+1\ge2a\) ; \(\dfrac{b^2}{a^2}+1\ge\dfrac{2b}{a}\) ; \(\dfrac{1}{b^2}+1\ge\dfrac{2}{b}\)

\(\Rightarrow a^2+\dfrac{b^2}{a^2}+\dfrac{1}{b^2}+3\ge a+\dfrac{b}{a}+\dfrac{1}{b}+a+\dfrac{b}{a}+\dfrac{1}{b}\ge a+\dfrac{b}{a}+\dfrac{1}{b}+3\sqrt[3]{\dfrac{ab}{ab}}\)

\(\Rightarrow a^2+\dfrac{b^2}{a^2}+\dfrac{1}{b^2}+3\ge a+\dfrac{b}{a}+\dfrac{1}{b}+3\)

\(\Rightarrow\) đpcm

Dấu "=" xảy ra khi \(a=b=1\)

Có cách khác ko ạ Nguyễn Việt Lâm Giáo viên

Sử dụng các bất đẳng thức quen thuộc ý ạ

\(3a^3+3b^3+3b^3+b^3\ge3\sqrt[3]{27a^3b^6}+b^3=9ab^2+b^3\ge9ab^2\)

Dấu "=" xảy ra khi \(a=b=0\)

\(\sum\dfrac{a}{\left(a^2+1\right)+2b+2}\le\sum\dfrac{a}{2\left(a+b+1\right)}=\dfrac{1}{2}\)

Nể''ss :D

\(vp=\frac{a\left(1+b\right)+b\left(1+a\right)}{\left(1+a\right)\left(1+b\right)}=\frac{2ab+a+b}{1+ab+a+b}\)

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

\(\ge\frac{a+b}{1+a+b}\)

Áp dụng BĐT cosi:

\(\left(2+a+b\right)\left(a+4b+ab\right)\ge3\sqrt[3]{2ab}\cdot3\sqrt[3]{4a^2b^2}=9\sqrt[3]{8a^3b^3}=9\cdot2ab=18ab\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}a=b=2\\a=4b=ab\end{matrix}\right.\left(\text{vô lí}\right)\)

Vậy dấu \("="\) ko xảy ra hay \(\left(2+a+b\right)\left(a+4b+ab\right)>18ab\)

\(\frac{a+b}{2}\le\sqrt{\frac{a^2+b^2}{2}}\)

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

\(\Leftrightarrow a+b\le\sqrt{2\left(a^2+b^2\right)}\)

\(\Leftrightarrow\left(a+b\right)^2\le2\left(a^2+b^2\right)\)

\(\Leftrightarrow2a^2+2b^2-a^2-2ab-b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\) ( luôn đúng )

Dấu "=" xảy ra \(\Leftrightarrow a=b\)