1)Từ đề bài:

`=>a^2+4b+4+b^2+4c+4+c^2+4a+4=0`

`<=>(a+2)^2+(b+2)^2+(c+2)^2=0`

`<=>a=b=c-2`

`ab+bc+ca=abc`

`<=>1/a+1/b+1/c=1`

`<=>(1/a+1/b+1/c)^2=1`

`<=>1/a^2+1/b^2+1/c^2+2/(ab)+2/(bc)+2/(ca)=1`

`<=>1/a^2+1/b^2+1/c^2=1-(2/(ab)+2/(bc)+2/(ca))`

`a+b+c=0`

Chia 2 vế cho `abc`

`=>1/(ab)+1/(bc)+1/(ca)=0`

`=>2/(ab)+2/(bc)+2/(ca)=0`

`=>1/a^2+1/b^2+1/c^2=1-0=1`

Hình như sai đề =)))

Min:

\(\left(a+b+c\right)^3=a^3+b^3+c^3+3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc\ge a^3+b^3+c^3\)

\(\Rightarrow a+b+c\ge\sqrt[3]{a^3+b^3+c^3}=\sqrt[3]{3}\)

\(\Rightarrow P=\dfrac{a}{7-3bc}+\dfrac{b}{7-3ca}+\dfrac{c}{7-3ab}\ge\dfrac{a}{7}+\dfrac{b}{7}+\dfrac{c}{7}=\dfrac{a+b+c}{7}\ge\dfrac{\sqrt[3]{3}}{7}\)

Dấu "=" xảy ra tại \(\left(a;b;c\right)=\left(0;0;\sqrt[3]{3}\right)\) và các hoán vị

Max:

\(\left(a^3+1+1\right)+\left(b^3+1+1\right)+\left(c^3+1+1\right)\ge3a+3b+3c\)

\(\Rightarrow a+b+c\le\dfrac{a^3+b^3+c^3+6}{3}=3\)

Khi đó:

\(7P=\dfrac{7a}{7-3bc}+\dfrac{7b}{7-3ca}+\dfrac{7c}{7-3ab}=\dfrac{a\left(7-3bc\right)+3abc}{7-3bc}+\dfrac{b\left(7-3ca\right)+3abc}{7-3ca}+\dfrac{c\left(7-3ab\right)+3abc}{7-3ab}\)

\(=a+b+c+\dfrac{3abc}{7-3bc}+\dfrac{3abc}{7-3ca}+\dfrac{3abc}{7-3ab}\)

Ta có:

\(7-3ab\ge\dfrac{7}{9}\left(a+b+c\right)^2-3ab=\dfrac{1}{9}\left[\dfrac{13}{2}\left(a-b\right)^2+\dfrac{1}{2}\left(a^2+b^2\right)+7c^2+14bc+14ca\right]\)

Do \(\dfrac{13}{2}\left(a-b\right)^2+\dfrac{1}{2}\left(a^2+b^2\right)\ge\dfrac{1}{2}\left(a^2+b^2\right)\ge ab\)

\(\Rightarrow7-3ab\ge\dfrac{1}{9}\left(ab+7c^2+14bc+14ca\right)\)

\(\Rightarrow\dfrac{3abc}{7-3ab}\le\dfrac{27abc}{ab+7c\left(c+2a+2b\right)}\le\dfrac{27abc}{36^2}\left(\dfrac{1^2}{ab}+\dfrac{35^2}{7c\left(c+2a+2b\right)}\right)\)

\(\Rightarrow\dfrac{3abc}{7-3ab}\le\dfrac{c}{48}+\dfrac{175}{48}.\dfrac{ab}{c+2a+2b}=\dfrac{c}{48}+\dfrac{175}{48}.\dfrac{ab}{\left(a+b+c\right)+\left(a+b\right)}\)

\(\Rightarrow\dfrac{3abc}{7-3ab}\le\dfrac{c}{48}+\dfrac{175}{48}.\dfrac{ab}{5^2}\left(\dfrac{3^2}{a+b+c}+\dfrac{2^2}{a+b}\right)\)

\(\Rightarrow\dfrac{3abc}{7-3ab}\le\dfrac{c}{48}+\dfrac{21}{16}.\dfrac{ab}{a+b+c}+\dfrac{7}{12}.\dfrac{ab}{a+b}\le\dfrac{c}{48}+\dfrac{21}{16}.\dfrac{ab}{a+b+c}+\dfrac{7}{48}.\dfrac{\left(a+b\right)^2}{a+b}\)

\(\Rightarrow\dfrac{3abc}{7-3ab}\le\dfrac{7a+7b+c}{48}+\dfrac{21}{16}.\dfrac{ab}{a+b+c}\)

Tương tự:

\(\dfrac{3abc}{7-3bc}\le\dfrac{a+7b+7c}{48}+\dfrac{21}{16}.\dfrac{bc}{a+b+c}\)

\(\dfrac{3abc}{7-3ca}\le\dfrac{7a+b+7c}{48}+\dfrac{21}{16}.\dfrac{ca}{a+b+c}\)

\(\Rightarrow7P\le\dfrac{21}{16}\left(a+b+c\right)+\dfrac{21}{16}\left(\dfrac{ab+bc+ca}{a+b+c}\right)\le\dfrac{21}{16}\left(a+b+c\right)+\dfrac{21}{48}.\dfrac{\left(a+b+c\right)^2}{a+b+c}\)

\(\Rightarrow7P\le\dfrac{7}{4}\left(a+b+c\right)\)

\(\Rightarrow P\le\dfrac{a+b+c}{4}\le\dfrac{3}{4}\)

Vậy \(P_{max}=\dfrac{3}{4}\) khi \(a=b=c=1\)

\(\dfrac{1}{c}+b^2c=ab\left(a+b+c\right)+b^2c=ab\left(a+c\right)+b^2\left(a+c\right)=b\left(a+b\right)\left(a+c\right)\)

\(\dfrac{1}{c}+a^2c=ab\left(a+b+c\right)+a^2c=a\left(a+b\right)\left(b+c\right)\)

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

\(\Leftrightarrow\left(1+b^2c^2\right)\left(1+a^2c^2\right)=c^2\left(a+b\right)^2ab\left(ab+bc+ac+c^2\right)\)\(=c^2\left(a+b\right)^2\left(a^2b^2+ab^2c+a^2bc+abc^2\right)\)\(=c^2\left(a+b\right)^2\left[a^2b^2+abc\left(a+b+c\right)\right]=c^2\left(a+b\right)^2\left(a^2b^2+1\right)\)

\(\Rightarrow\dfrac{\left(1+b^2c^2\right)\left(1+a^2c^2\right)}{c^2\left(a^2b^2+1\right)}=\left(a+b\right)^2\)

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

tịt

a) Áp dụng bất đẳng thức Schur với \(r=1\)

\(\Rightarrow a^3+b^3+c^3+3abc\ge a^2b+ab^2+b^2c+bc^2+c^2a+ca^2\)

\(\Rightarrow3abc\ge a^2b+ca^2-a^3+ab^2+b^2c-b^3+c^2a+bc^2-c^3\)

\(\Rightarrow3abc\ge a^2\left(b+c-a\right)+b^2\left(a+c-b\right)+c^2\left(a+b-c\right)\) ( đpcm )

Dấu " = " xảy ra khi \(a=b=c\)

b) Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\dfrac{a^3}{b^2}+b+b\ge3\sqrt[3]{\dfrac{a^3}{b^2}.b^2}=3a\)

Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{b^3}{c^2}+c+c\ge3b\\\dfrac{c^3}{a^2}+a+a\ge3c\end{matrix}\right.\)

\(\Rightarrow\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}+2\left(a+b+c\right)\ge3\left(a+b+c\right)\)

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

Dấu " = " xảy ra khi \(a=b=c\)

c) Ta có \(abc=ab+bc+ca\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0

\(\Rightarrow\dfrac{1}{a+2b+3c}=\dfrac{1}{a+c+2\left(b+c\right)}\le\dfrac{1}{4}\left[\dfrac{1}{a+c}+\dfrac{1}{2\left(b+c\right)}\right]\)

Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+2c+3a}\le\dfrac{1}{4}\left[\dfrac{1}{a+b}+\dfrac{1}{2\left(a+c\right)}\right]\\\dfrac{1}{c+2a+3b}\le\dfrac{1}{4}\left[\dfrac{1}{b+c}+\dfrac{1}{2\left(a+b\right)}\right]\end{matrix}\right.\)

\(\Rightarrow VT\le\dfrac{1}{4}\left[\dfrac{3}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\right]\)

\(\Rightarrow VT\le\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\) ( 1 )

Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0

\(\Rightarrow\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+c}\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\\\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\end{matrix}\right.\)

\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\right]\)

\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\right]\)

\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{16}\) ( 2 )

Từ ( 1 ) và ( 2 )

\(\Rightarrow VT\le\dfrac{3}{16}\)

\(\Rightarrow\dfrac{1}{a+2b+3c}+\dfrac{1}{b+2c+3a}+\dfrac{1}{c+2a+3b}\le\dfrac{3}{16}\) ( đpcm )

Câu 2 b sos là ra :D

\(BĐT\Leftrightarrow\Sigma_{cyc}\left(\frac{a^3-ab^2}{b^2}\right)\ge0\Leftrightarrow\Sigma_{cyc}\left(\frac{a\left(a-b\right)\left(a+b\right)}{b^2}-2\left(a-b\right)\right)\ge0\)

\(\Leftrightarrow\Sigma_{cyc}\left(a-b\right)\left(\frac{a^2-b^2+ab-b^2}{b^2}\right)\ge0\)\(\Leftrightarrow\Sigma_{cyc}\frac{\left(a-b\right)^2\left(a+2b\right)}{b^2}\ge0\)

BĐT cuối cùng đúng nên ta có Q.E.D.

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

Câu 3 có cách khác ạ,nhưng mà sao em thấy nó sai sai,vì dấu "=" không xảy ra!Mong mọi người check giúp ạ!

gt<=> \(abc=ab+bc+ca\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\) (chia hai vế cho abc>0)

Dễ chứng minh \(\frac{1}{36}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{1}{k}\right)\ge\frac{1}{a+b+c+d+e+k}\)

Áp dụng vào,ta có: \(VT\le\frac{6}{36}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{6}< \frac{3}{16}?!?\)