Bạn xem lại đề. Cho $a=b=c=1$ thì BĐT sai.

bạn chép sai đề ak

ko cần đâu mình ra rồi

BĐT 1 sai ngay với \(a=\sqrt{0,1},b=\sqrt{0,2},c=\sqrt{2,7}\)

BĐT 2 tương đương với đi chứng minh \(a^4b^4+b^4c^4+c^4a^4\geq 3a^2b^2c^2\)

Áp dụng BĐT AM-GM: \(a^4b^4+b^4c^4\geq 2a^2b^4c^2\)

Tương tự \(b^4c^4+c^4a^4\geq 2b^2c^4a^2,a^4b^4+c^4a^4\geq 2a^4b^2c^2\)

Cộng theo vế và rút gọn:

\(\Rightarrow a^4b^4+b^4c^4+c^4a^4\geq a^2b^2c^2(a^2+b^2+c^2)=3a^2b^2c^2\)

Do đó ta có đpcm. Dấu $=$ xảy ra khi $a=b=c=1$

thì ra cái đầu sai nghĩ mãi ko ra, đại ca thông minh thật :v

\(VT=\frac{a^3}{b^2+8}+\frac{b^3}{c^2+8}+\frac{c^3}{a^2+8}\)

\(=\frac{a^3}{b^2+ab+bc+ca}+\frac{b^3}{c^2+ab+bc+ca}+\frac{c^3}{a^2+ab+bc+ca}\)

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

Áp dụng BĐT Cô si ta có :

\(\left\{{}\begin{matrix}\frac{a^3}{\left(a+b\right)\left(b+c\right)}+\frac{a+b}{8}+\frac{b+c}{8}\ge\frac{3a}{4}\\\frac{b^3}{\left(b+c\right)\left(c+a\right)}+\frac{b+c}{8}+\frac{c+a}{8}\ge\frac{3b}{4}\\\frac{c^3}{\left(c+a\right)\left(a+b\right)}+\frac{c+a}{8}+\frac{a+b}{8}\ge\frac{3c}{4}\end{matrix}\right.\)

\(\Rightarrow\frac{a^3}{\left(a+b\right)\left(b+c\right)}+\frac{b^3}{\left(b+c\right)\left(c+a\right)}+\frac{c^3}{\left(c+a\right)\left(a+b\right)}\ge\frac{a+b+c}{4}\ge\frac{\sqrt{3\left(ab+bc+ca\right)}}{4}=\frac{3}{4}\)

Vậy BĐT được chứng minh . Dấu = xảy ra khi \(a=b=c=1\)

Lời giải:

Ta thấy:

\(\text{VT}=(a+\frac{ca}{a+b})+(b+\frac{ab}{b+c})+(c+\frac{bc}{c+a})\)

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

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

\(\geq (a+b+c).\frac{(a+b+c)^2}{a^2+ab+b^2+bc+c^2+ac}=\frac{(a+b+c)^3}{a^2+b^2+c^2+ab+bc+ac}\) (theo BĐT Cauchy-Schwarz)

Có:

$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)=a^2+b^2+c^2+2$

$\Rightarrow a+b+c=\sqrt{a^2+b^2+c^2+2}=\sqrt{t+2}$ với $t=a^2+b^2+c^2$

Do đó:

$\text{VT}\geq \frac{\sqrt{(t+2)^3}}{t+1}$ \(=\sqrt{\frac{(t+2)^3}{(t+1)^2}}\)

Áp dụng BĐT AM-GM:

\((t+2)^3=\left(\frac{t+1}{2}+\frac{t+1}{2}+1\right)^3\geq 27.\frac{(t+1)^2}{4}\)

\(\Rightarrow \text{VT}=\sqrt{\frac{(t+2)^3}{(t+1)^2}}\geq \sqrt{\frac{27}{4}}=\frac{3\sqrt{3}}{2}\) (đpcm)

Dấu "=" xảy ra khi $a=b=c=\frac{1}{\sqrt{3}}$

Mày chỉ tao SOS đi :((

A=\(\frac{a}{3a+b+c}+\frac{b}{3b+a+c}+\frac{c}{3c+a+b}\)

=>\(\frac{3}{2}\)-A=\(\frac{1}{2}-\frac{a}{3a+b+c}+\frac{1}{2}-\frac{b}{3b+a+c}+\frac{1}{2}-\frac{c}{3c+a+b}\)

<=>\(\frac{3}{2}\)-A=\(\left(a+b+c\right)\left(\frac{1}{6a+2b+2c}+\frac{1}{6b+2a+2c}+\frac{1}{6c+2a+2b}\right)\)

ta lại có

\(\left(a+b+c\right)\left(\frac{1}{6a+2b+2c}+\frac{1}{6b+2a+2c}+\frac{1}{6c+2a+2b}\right)\ge\left(a+b+c\right)\left(\frac{\left(1+1+1\right)^2}{6a+2b+2c+6b+2a+2c+6c+2a+2b}\right)=\frac{9}{10}\)<=>\(\frac{3}{2}-\)A\(\ge\frac{9}{10}\)<=>A\(\le\frac{3}{2}-\frac{9}{10}=\frac{3}{5}\)

dấu "=" xảy ra <=>a=b=c

Lời giải:

Ta có:
\(\text{VT}=\frac{a^3}{b^2+3}+\frac{b^3}{c^2+3}+\frac{c^3}{a^2+3}=\frac{a^3}{b^2+ab+bc+ac}+\frac{b^3}{c^2+ab+bc+ac}+\frac{c^3}{a^2+ab+bc+ac}\)

\(=\frac{a^3}{(b+a)(b+c)}+\frac{b^3}{(c+a)(c+b)}+\frac{c^3}{(a+b)(a+c)}\)

Áp dụng BĐT AM-GM:
\(\frac{a^3}{(b+a)(b+c)}+\frac{b+a}{8}+\frac{b+c}{8}\geq 3\sqrt[3]{\frac{a^3}{8.8}}=\frac{3a}{4}\)

\(\frac{b^3}{(c+a)(c+b)}+\frac{c+a}{8}+\frac{c+b}{8}\geq \frac{3b}{4}\)

\(\frac{c^3}{(a+b)(a+c)}+\frac{a+b}{8}+\frac{a+c}{8}\geq \frac{3c}{4}\)

Cộng theo vế và rút gọn thu được:

\(\text{VT}\geq \frac{a+b+c}{4}\)

Tiếp tục áp dụng BĐT AM-GM: \((a+b+c)^2\geq 3(ab+bc+ac)=9\Rightarrow a+b+c\geq 3\)

Do đó: \(\text{VT}\geq \frac{3}{4}\) (đpcm)

Dấu "=" xảy ra khi $a=b=c=1$