Ta có : \(\frac{3+a^2}{b+c}+\frac{3+b^2}{c+a}+\frac{3+c^2}{a+b}=\frac{3}{b+c}+\frac{3}{c+a}+\frac{3}{a+b}+\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\)

Ta cầm chứng minh : \(\hept{\begin{cases}\frac{3}{a+b}+\frac{3}{a+c}+\frac{3}{b+c}\ge\frac{9}{2}\left(1\right)\\\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\ge\frac{3}{2}\left(2\right)\end{cases}}\)

Ta có bđt (1) \(\Leftrightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}\ge\frac{9}{2}\)

\(\Leftrightarrow2\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)\ge9\)

\(\Leftrightarrow\left[\left(a+b\right)+\left(b+c\right)+\left(a+c\right)\right]\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)\ge9\)

Áp dụng bđt AM GM ta có :

\(\hept{\begin{cases}\left(a+b\right)+\left(b+c\right)+\left(a+c\right)\ge3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\\\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\ge\frac{3}{\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(a+c\right)}}\end{cases}}\)

Nhân vế với vế ta được đpcm ; Vậy bđt (1) đc chứng minh

Ta có \(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}=\frac{3}{2}\) 

Vậy bđt (2) đc chứng minh

Do 2 bất đẳng thức dước chứng minh

\(\Rightarrow\frac{3+a^2}{b+c}+\frac{3+b^2}{a+c}+\frac{3+c^2}{a+b}\ge\frac{3}{2}+\frac{9}{2}=6\) (ĐPCM)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)

tách ra mình làm cho. để cả đống này k làm đc đâu

ý a, áp dụng BĐT cô si có 

   a + b >= căn ab     dấu = xay ra a=b

b + c >= căn bc         dau = xay ra khi b=c

c+a >= căn ac           dau = xay ra khi a=c

công tung ve vao. rut gon ta dc điều phải chung minh

sử dụng bdt bunhiacopxki có đc ko bn

\(a^2\sqrt{a}+b^2\sqrt{b}+c^2\sqrt{c}+\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\)

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

\(\ge2a+2b+2c\ge6\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=6\)

\(P=\frac{b+c+5}{1+a}+\frac{c+a+4}{2+b}+\frac{a+b+3}{3+c}\)

\(\Rightarrow P+3=\frac{b+c+5}{1+a}+1+\frac{c+a+4}{2+b}+1+\frac{a+b+3}{3+c}+1\)

\(\Rightarrow P+3=\frac{a+b+c+6}{1+a}+\frac{a+b+c+6}{2+b}+\frac{a+b+c+6}{3+c}\)

\(\Rightarrow P+3=\frac{12}{1+a}+\frac{12}{2+b}+\frac{12}{3+c}\ge\frac{12.9}{6+a+b+c}=9\)

\(\Rightarrow P\ge6\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=3\\b=2\\c=1\end{matrix}\right.\)

này là dấu gì @Nguyễn Việt Lâm

Đặt vế trái là P

\(P=\frac{b+c+1}{a+5}+1+\frac{c+a+4}{b+2}+1+\frac{a+b+3}{c+3}+1-3\)

\(P=\frac{a+b+c+6}{a+5}+\frac{a+b+c+6}{b+2}+\frac{a+b+c+6}{c+3}-3\)

\(P=\frac{12}{a+5}+\frac{12}{b+2}+\frac{12}{c+3}-3\)

\(P\ge\frac{12.9}{a+b+c+10}-3=\frac{12.9}{6+10}-3=\frac{15}{4}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=\frac{1}{3}\\b=\frac{10}{3}\\c=\frac{7}{3}\end{matrix}\right.\)

Đề bài sai :)

Áp dụng BĐT cô si với hai số không âm, Ta có: 

\(\left(a+b+c\right)^2=1\ge4a\left(b+c\right)\)

\(\Leftrightarrow b+c\ge4a\left(b+c\right)^2\)

Mà \(\left(b+c\right)^2\ge4bc\forall b,c\ge0\)

\(\Rightarrow b+c\ge16abc\)

Dấu "=" xảy ra khi: 

\(\hept{\begin{cases}a+b+c=1\\b=c\\a=b+c\end{cases}}\Rightarrow\hept{\begin{cases}a=\frac{1}{2}\\b=c=\frac{1}{4}\end{cases}}\)

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

\(\Leftrightarrow3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge9\)

\(\Leftrightarrow\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge6\)

Áp dụng BĐT Cô si với 2 số dương ta có: 

\(\frac{a}{b}+\frac{b}{a}\ge2,\frac{b}{c}+\frac{c}{b}\ge2,\frac{c}{a}+\frac{a}{c}\ge2\)

\(\Leftrightarrow\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge6\)(đúng) 

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)(do a+b+c=1)

Ta có: \(a< a+b\left(a,b>0\right)\Rightarrow\frac{a}{a+b}< 1\)

Có: \(\frac{a}{a+b}=\sqrt{\frac{a}{a+b}}.\sqrt{\frac{a}{a+b}}\)

Lại có: \(\frac{a}{b+a}< 1\Leftrightarrow\sqrt{\frac{a}{b+a}}< 1\Rightarrow\sqrt{\frac{a}{a+b}}.\sqrt{\frac{a}{a+b}}< \sqrt{\frac{a}{a+b}}\Rightarrow\frac{a}{a+b}< \sqrt{\frac{a}{a+b}}\)

Chứng minh tương tự ta có:

\(\frac{b}{b+c}< \sqrt{\frac{b}{b+c}}\)

\(\frac{c}{c+a}< \sqrt{\frac{c}{c+a}}\)

\(\Rightarrow\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< \sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\)

                                                                                               đpcm

Sai thì thôi nhé~

Mới lp 8