cau b . ta co 

a⁴+b^{4\(\ge\frac{\left(a^2+b^2\right)^2}{2}\)}\(\ge\)\(\frac{\frac{1}{16}}{2}\)=1/32

câu a đề phải là 12ab 

Dùng BĐT cô si 

\(ab\ge2\sqrt{ab}\)

\(9+ab\ge2.3\sqrt{ab}\)

\(\Rightarrow\left(a+b\right)\left(9+ab\right)\ge12ab\)

Sửa đề: \(CMR:a+b\ge\frac{12ab}{9+ab}\)

 Áp dụng BĐT Cô-si cho 2 số không âm ta có: 

      \(a+b\ge2\sqrt{ab}\)

       \(9+ab\ge6\sqrt{ab}\)

\(\Rightarrow\left(a+b\right)\left(9+ab\right)\ge12ab\)

\(\Rightarrow a+b\ge\frac{12ab}{9+ab}\)

đề đúng: \(a,b,c>0\)

chuẩn hoá: \(a+b+c=3\)

\(\frac{1}{a^2+ab}+\frac{a}{2}+\frac{a+b}{4}\ge\frac{3}{2}\)\(\Leftrightarrow\)\(\frac{1}{a^2+ab}\ge\frac{3}{2}-\frac{3}{4}a-\frac{1}{4}b\)

tương tự \(\Rightarrow\)\(\Sigma\frac{1}{a^2+ab}\ge\frac{9}{2}-\left(a+b+c\right)=\frac{3}{2}=\frac{27}{2\left(a+b+c\right)^2}\)

dấu "=" xảy ra khi \(a=b=c=1\)

chưa học chuẩn hoá thì dùng cách này: 

gia su: \(a+b+c=3k>0\)

\(\frac{1}{a^2+ab}+\frac{a}{2k^3}+\frac{a+b}{4k^3}\ge\frac{3}{2k^2}\)\(\Leftrightarrow\)\(\frac{1}{a^2+ab}\ge\frac{3}{2k^2}-\frac{3}{4k^3}a-\frac{1}{4k^3}b\)

\(\Rightarrow\)\(\Sigma\frac{1}{a^2+ab}\ge\frac{9}{2k^2}-\frac{a+b+c}{4k^3}=\frac{3}{2k^2}=\frac{27}{2\left(a+b+c\right)^2}\)

dấu "=" xảy ra khi \(a=b=c=k\)

Có cách khác không thấy áp đặt ở cách 2 quá còn cách chuẩn hóa thì cảm giác không ổn

\(\frac{1}{a^2+ab}\ge\frac{2}{\frac{1}{4}\left(3a+b\right)^2}\)

\(\Rightarrow\Sigma_{cyc}\frac{1}{a^2+ab}\ge\Sigma_{cyc}\frac{8}{\left(3a+b\right)^2}\ge8\frac{\left(\frac{1}{3a+b}+\frac{1}{3b+c}+\frac{1}{3c+a}\right)^2}{3}\ge\frac{8\frac{81}{16\left(a+b+c\right)^2}}{3}=\frac{27}{2\left(a+b+c\right)^2}\)

Áp dụng cô si

\(\hept{\begin{cases}\frac{1}{a}+\frac{1}{b}\ge2\sqrt{\frac{1}{ab}}\\\frac{1}{c}+\frac{1}{b}\ge2\sqrt{\frac{1}{cb}}\\\frac{1}{a}+\frac{1}{c}\ge2\sqrt{\frac{1}{ac}}\end{cases}}\)\(\Rightarrow\frac{1}{c}+\frac{1}{b}+\frac{1}{a}\ge\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ac}}\)

\("="\Leftrightarrow a=b=c=0\)

\(\hept{\begin{cases}\sqrt{x}\le\frac{x+1}{2}\\\sqrt{y-1}\le\frac{y-1+1}{2}\\\sqrt{z-2}\le\frac{z-2+1}{2}\end{cases}}\)\(\Rightarrow\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}\le\frac{x+1+y-1+1+z-2+1}{2}\)

\(\Leftrightarrow\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}\le\frac{x+y+z}{2}\)

\("="\Leftrightarrow\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}\)

Sửa ĐK của c) : a, b, c > 0

Áp dụng bất đẳng thức Cauchy ta có :

\(\frac{1}{a}+\frac{1}{b}\ge2\sqrt{\frac{1}{ab}}=\frac{2}{\sqrt{ab}}\)

\(\frac{1}{b}+\frac{1}{c}\ge2\sqrt{\frac{1}{bc}}=\frac{2}{\sqrt{bc}}\)

\(\frac{1}{c}+\frac{1}{a}\ge2\sqrt{\frac{1}{ca}}=\frac{2}{\sqrt{ca}}\)

Cộng các vế tương ứng

=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\ge\frac{2}{\sqrt{ab}}+\frac{2}{\sqrt{bc}}+\frac{2}{\sqrt{ca}}\)

=> \(2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge2\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\right)\)

=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\)

=> đpcm

Đẳng thức xảy ra khi a = b = c

c) Cách khác: Áp dụng bổ đề: \(x^2+y^2+z^2\ge xy+yz+zx\forall x,y,z>0\)

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

\(=\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\)

Dấu "=" xảy ra khi \(a=b=c>0\)

Nhân 2 lên rồi cô-si

áp dụng bất đẳng thức \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)ta có:

\(\left(\frac{ac}{b}+\frac{bc}{a}+\frac{ca}{b}\right)^2\ge3\left(a^2+b^2+c^2\right)=3\)

\(\Rightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge\sqrt{3}\left(Q.E.D\right)\)

\(a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)\ge\left(a+b\right)\left(2ab-ab\right)=ab\left(a+b\right)\)

\(\frac{1}{a^3+b^3+1}\le\frac{1}{ab\left(a+b\right)+1}=\frac{abc}{ab\left(a+b\right)+abc}=\frac{abc}{ab\left(a+b+c\right)}=\frac{c}{a+b+c}\)

Tương tự \(\frac{1}{b^3+c^3+1}\le\frac{a}{a+b+c}\); \(\frac{1}{a^3+c^3+1}\le\frac{b}{a+b+c}\)

Cộng vế với vế:

\(\sum\frac{1}{a^3+b^3+1}\le\frac{a+b+c}{a+b+c}=1\)(đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)