Với mọi a;b dương ta có:

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

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

Do đó:

\(A\le\sum\dfrac{ab}{ab\left(a^2+b^2\right)+ab}+2020=\sum\dfrac{1}{a^2+b^2+1}+2020\)

Đặt \(\left(a^2;b^2;c^2\right)=\left(x^3;y^3;z^3\right)\Rightarrow xyz=1\)

\(\Rightarrow A\le\sum\dfrac{1}{x^3+y^3+1}+2020\le\sum\dfrac{1}{xy\left(x+y\right)+1}+2020\)

\(A\le\sum\dfrac{xyz}{xy\left(x+y\right)+xyz}+2020=\sum\dfrac{z}{x+y+z}+2020=1+2020=2021\)

Dấu "=" xảy ra khi \(a=b=c=1\)

\(\dfrac{a}{1+a}+\dfrac{b^3}{1+b}=1\Rightarrow\dfrac{a}{1+a}=1-\dfrac{b^3}{1+b}=\dfrac{1+b-b^3}{1+b}\)

\(\Rightarrow a\left(1+b\right)=\left(1+a\right)\left(1+b-b^3\right)\)

\(\Rightarrow a\left(1+b\right)=1+b-b^3+a\left(1+b-b^3\right)\)

\(\Rightarrow ab^3=1+b-b^3=1+3.\dfrac{1}{\sqrt[]{3}}.\dfrac{1}{\sqrt[]{3}}b-b^3\le1+\left(\dfrac{1}{\sqrt[]{3}}\right)^3+\left(\dfrac{1}{\sqrt[]{3}}\right)^3+b^3-b^3=\dfrac{9+2\sqrt[]{3}}{9}\)

\(P_{max}=\dfrac{9+2\sqrt[]{3}}{9}\) khi \(\left(a;b\right)=\left(2+3\sqrt[]{3};\dfrac{1}{\sqrt[]{3}}\right)\)

Lời giải:

$N=a(b+3c)+5bc=(1-b-c)(b+3c)+5bc$

$=b+3c-b^2-3c^2+bc$

$-N=b^2+3c^2-bc-b-3c$

$-2N=2b^2+6c^2-2bc-2b-6c$

$\geq b^2+5c^2-2b-6c$

$=(b+c-1)^2+(2c-1)^2-2bc-2$

$\geq -2(bc+1)$

Mà $bc\leq \frac{(b+c)^2}{4}\leq \frac{1}{4}$

$\Rightarrow bc+1\leq \frac{5}{4}$

$\Rightarrow -2(bc+1)\geq \frac{-10}{4}$
$\Rightarrow -2N\geq \frac{-10}{4}$

$\Rightarrow N\leq \frac{5}{4}$

Vậy $N_{\max}=\frac{5}{4}$ khi $(a,b,c)=(0,\frac{1}{2}, \frac{1}{2})$

Theo tớ là tìm Min chứ nhỉ ??

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

\(A=\dfrac{a^3+b^3}{a^3b^3}=\dfrac{\left(a+b\right)\left(a^2+b^2-ab\right)}{a^3b^3}=\dfrac{\left(a+b\right)ab\left(a+b\right)}{a^3b^3}=\dfrac{\left(a+b\right)^2}{a^2b^2}\)

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

Ta có: \(a^2+b^2-ab>0;\forall a;b\ne0\Rightarrow\dfrac{\left(a+b\right)^2}{a^2+b^2-ab}\ge0\)

\(\dfrac{\left(a+b\right)^2}{a^2+b^2-ab}=\dfrac{a^2+b^2+2ab}{a^2+b^2-ab}=\dfrac{4\left(a^2+b^2-ab\right)-3\left(a^2+b^2-2ab\right)}{a^2+b^2-ab}=4-\dfrac{3\left(a-b\right)^2}{a^2+b^2-ab}\le4\)

\(\Rightarrow0\le\dfrac{\left(a+b\right)^2}{a^2+b^2-ab}\le4\)

\(\Rightarrow A\le16\)

\(A_{max}=16\) khi \(a=b=\dfrac{1}{2}\)

ĐK : a;b;c > 0 

Ta có : \(ab+bc+ac=1\) \(\Leftrightarrow c\left(a+b\right)=1-ab\Leftrightarrow c=\dfrac{1-ab}{a+b}\)

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

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

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

Suy ra : \(A=\dfrac{a}{a^2+1}+\dfrac{b}{b^2+1}-\dfrac{1}{c^2+1}=\dfrac{\left(a+b\right)\left(ab+1-a-b\right)}{\left(a^2+1\right)\left(b^2+1\right)}=\dfrac{\left(a+b\right)\left(1-a\right)\left(1-b\right)}{\left(a^2+1\right)\left(b^2+1\right)}\)

AD BĐT Cauchy ta được :  \(\left(a+b\right)\left[\left(1-a\right)\left(1-b\right)\right]\le\dfrac{\left[a+b+\left(1-a\right)\left(1-b\right)\right]^2}{4}=\dfrac{\left(1+ab\right)^2}{4}\)

\(\left(a^2+1\right)\left(b^2+1\right)\ge\left(ab+1\right)^2\)  ( theo BCS )

Suy ra : \(A\le\dfrac{1}{4}\)

Theo Cô si:

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

\(\Rightarrow2A\le\left(\frac{1}{2}\right)^2=\frac{1}{4}\Rightarrow A\le\frac{\left(\frac{1}{4}\right)}{2}=\frac{1}{8}\)

Vậy \(A_{max}=\frac{1}{8}\Leftrightarrow a=b=\frac{1}{2}\)

Cosi cho 2 số dương nha @tth 

hey tth có cho đk: a,b >= 0?!?

2

Đặt OA = a ; OB = b ; OC = c . Khi đó : 

\(OA+OB+OC+AB+BC+AC=a+b+c+\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{a^2+c^2}\)

AD BĐT Cauchy ta được : \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{a^2+c^2}\ge\sqrt{2}\left(a+b+c\right)\)

Suy ra : l \(\ge\left(\sqrt{2}+1\right)\left(a+b+c\right)\ge\left(\sqrt{2}+1\right)3\sqrt[3]{abc}\)

Có : \(V=V_{OABC}=\dfrac{abc}{6}\)  . Suy ra :   \(l\ge3\left(\sqrt{2}+1\right)\sqrt[3]{6V}\Leftrightarrow V\le\dfrac{l^3}{27\left(\sqrt{2}+1\right)^3.6}=\dfrac{l^3}{162\left(\sqrt{2}+1\right)^3}\)

" = " \(\Leftrightarrow a=b=c\) = \(\dfrac{l\left(\sqrt{2}-1\right)}{3}\)

\(\frac{ab}{c+1}=\frac{ab}{a+c+b+c}\le\frac{ab}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)=\frac{1}{4}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)\)

\(M\le\frac{1}{4}\left[\frac{ab}{a+c}+\frac{ab}{b+c}+\frac{bc}{a+b}+\frac{bc}{c+a}+\frac{ca}{a+b}+\frac{ca}{b+c}\right]\)

\(=\frac{1}{4}\left[\frac{b\left(a+c\right)}{a+c}+\frac{a\left(b+c\right)}{b+c}+\frac{c\left(a+b\right)}{a+b}\right]=\frac{1}{4}\left(a+b+c\right)=\frac{1}{4}\)

 chịu thôi chị ơi!

Ai trả lời câu này được bái luôn thành sư phụ!!!!!

cảm ơn nha, giờ chị giải được rồi