\(A=ab+\dfrac{1}{ab}+2=ab+\dfrac{1}{16ab}+\dfrac{15}{16}ab+2\)

\(A\ge2\sqrt{\dfrac{ab}{16ab}}+\dfrac{15}{4\left(a+b\right)^2}+2=\dfrac{25}{4}\)

Dấu "=" xảy ra khi \(a=b=\dfrac{1}{2}\)

`A=(a+1/b)(b+1/a)`

`=ab+1+1+1/(ab)`

`=2+ab+1/(16ab)+15/(16ab)`

Áp dụng cosi

`=>ab+1/(16ab)>=1/2`

`ab<=(a+b)^2/4=1/4`

`=>16ab<=4`

`=>15/(16ab)>=15/4`

`=>A>=15/4+1/2+2=25/4`

Dấu "=" xảy ra khi `a=b=1/2`

\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)

\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)

\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)

\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)

\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)

\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)

\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

p \(\ge\)\(\frac{4}{a^2+b^2+2\left(a+b\right)}\) +\(\sqrt{\left(1+ab\right)^2}\) (bunhia và cosi)

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

do \(a+b=ab\le\frac{\left(a+b\right)^2}{4}\Rightarrow a+b\ge4\)

dạt a+b = t thì t>=4

cần tìm min \(\frac{4}{t^2}+t+1=\frac{4}{t^2}+\frac{t}{16}+\frac{t}{16}+\frac{7t}{8}+1\)

                                      \(\ge3.\sqrt[3]{\frac{4}{t^2}.\frac{t}{16}.\frac{t}{16}}+\frac{7.4}{8}+1=\frac{21}{4}\)

dau = xay ra khi a=b=2

24+t−94(∗)
Xét hàm (∗) được: MinF(t)=F(23)=−19
⇒MinP=MinF(t)=−19.dấu "=" xảy ra khi a=b=c=13

\(Q=\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+bc}}\ge\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+\dfrac{1}{4}\left(b+c\right)^2}}=\dfrac{2}{3}\sum\dfrac{\left(a+b\right)^2}{b+c}\)

\(Q\ge\dfrac{2}{3}.\dfrac{\left(a+b+b+c+c+a\right)^2}{a+b+b+c+c+a}=\dfrac{4}{3}\left(a+b+c\right)=\dfrac{4}{3}\)

Lời giải:
Vì $abc=1$ nên:

\((a+bc)(b+ac)(c+ab)=a(a+bc)b(b+ac)c(c+ab)=(a^2+1)(b^2+1)(c^2+1)\)

Áp dụng BĐT Bunhiacopxky:

\((a^2+1)(1+b^2)\geq (a+b)^2; (a^2+1)(1+c^2)\geq (a+c)^2; (b^2+1)(1+c^2)\geq (b+c)^2\)

Nhân theo vế và thu gọn:

\(\Rightarrow (a^2+1)(b^2+1)(c^2+1)\geq (a+b)(b+c)(c+a)\)

Lại có: Theo BĐT AM-GM thì:

\((a+b)(b+c)(c+a)=(ab+bc+ac)(a+b+c)-abc\)

\(\geq (ab+bc+ac)(a+b+c)-\frac{(a+b+c)(ab+bc+ac)}{9}=\frac{8(a+b+c)(ab+bc+ac)}{9}(*)\) (đây là BĐT khá quen thuộc rồi)

Do đó:

\(P=\frac{(a+bc)(b+ca)(c+ab)}{ab+bc+ac}+\frac{1}{a+b+c}=\frac{(a^2+1)(b^2+1)(c^2+1)}{ab+bc+ac}+\frac{1}{a+b+c}\geq \frac{(a+b)(b+c)(c+a)}{ab+bc+ac}+\frac{1}{a+b+c}\)

\(P\geq \frac{7(a+b)(b+c)(c+a)}{8(ab+bc+ac)}+\frac{(a+b)(b+c)(c+a)}{8(ab+bc+ac)}+\frac{1}{a+b+c}\)

Áp dụng BĐT (*) và AM-GM:

\(\frac{7(a+b)(b+c)(c+a)}{8(ab+bc+ac)}\geq 7.\frac{\frac{8}{9}(a+b+c)(ab+bc+ac)}{8(ab+bc+ac)}=\frac{7}{9}(a+b+c)\geq \frac{7}{9}.3\sqrt[3]{abc}=\frac{7}{3}\)

\(\frac{(a+b)(b+c)(c+a)}{8(ab+bc+ac)}+\frac{1}{a+b+c}\geq 2\sqrt{\frac{(a+b)(b+c)(c+a)}{8(ab+bc+ac)(a+b+c)}}\geq 2\sqrt{\frac{\frac{8}{9}(a+b+c)(ab+bc+ac)}{8(a+b+c)(ab+bc+ac)}}=\frac{2}{3}\)

\(\Rightarrow P\geq \frac{7}{3}+\frac{2}{3}=3\)

Vậy $P_{\min}=3$

\(\left(a+bc\right)\left(b+ca\right)\left(c+ab\right)\)

\(=a^2+b^2+c^2+a^2b^2+b^2c^2+c^2a^2+1+1\)

\(=a^2+b^2+c^2+a^2b^2+b^2c^2+c^2a^2+1+1+1-1\)

Áp dụng BĐT AM-GM ta có:

\(\left(a+bc\right)\left(b+ca\right)\left(c+ab\right)\ge a^2+b^2+c^2+2ab+2bc+2ac-1=\left(a+b+c\right)^2-1\)\(\Rightarrow P\ge\frac{\left(a+b+c\right)^2-1}{ab+bc+ca}+\frac{1}{a+b+c}\)

Dấu " = " xảy ra <=> ...

Ta có: \(\frac{1}{3}.\left(a+b+c\right)^2\ge ab+bc+ca\)( BĐT quen thuộc tự c/m)

\(\Rightarrow P\ge\frac{\left(a+b+c\right)^2-1}{ab+bc+ca}+\frac{1}{a+b+c}\ge\frac{\left(a+b+c\right)^2}{\frac{1}{3}\left(a+b+c\right)^2}-\frac{1}{\frac{1}{3}\left(a+b+c\right)}+\frac{1}{a+b+c}\)\(=3+\frac{a+b+c-3}{\left(a+b+c\right)^2}\)

Ta có: \(abc=1\Leftrightarrow\sqrt[3]{abc}=1\le\frac{a+b+c}{3}\left(AM-GM\right)\)

\(\Rightarrow a+b+c\ge3\)

Dấu " = " xảy ra <=> ...

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

Dấu " = " xảy ra <=> a=b=c=1

KL:...........

alibaba nguyễn giúp em với  WTFシSnow 

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