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a + b + c= 1 \(\Rightarrow\)1 - a = b + c > 0
Tương tự : 1 - b > 0 ; 1 - c > 0
Mà 1 + a = 1 + ( 1 - b - c ) = ( 1- b ) + ( 1 - c ) \(\ge\)\(2\sqrt{\left(1-b\right)\left(1-c\right)}\)
Tương tự : \(1+b\ge2\sqrt{\left(1-a\right)\left(1-c\right)}\); \(1+c\ge2\sqrt{\left(1-a\right)\left(1-b\right)}\)
\(\Rightarrow\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge8\sqrt{\left(1-a\right)^2\left(1-b\right)^2\left(1-c\right)^2}=8\left(1-a\right)\left(1-b\right)\left(1-c\right)\)
\(\Rightarrow A=\frac{\left(1+a\right)\left(1+b\right)\left(1+c\right)}{\left(1-a\right)\left(1-b\right)\left(1-c\right)}\ge8\)
Dấu " = : xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)
Vậy GTNN của A là 8 \(\Leftrightarrow a=b=c=\frac{1}{3}\)
Cách khác:
\(A=\frac{\left[\left(a+b\right)+\left(a+c\right)\right]\left[\left(b+c\right)+\left(b+a\right)\right]\left[\left(c+a\right)+\left(c+b\right)\right]}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Áp dụng BĐT Cô si cho 2 số ta được:
\(A\ge\frac{8\sqrt{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=8\)
"=" <=> a = b = c = 1/3
Kết luận..
\(Q=\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+bc}}\ge\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+\dfrac{1}{4}\left(b+c\right)^2}}=\dfrac{2}{3}\sum\dfrac{\left(a+b\right)^2}{b+c}\)
\(Q\ge\dfrac{2}{3}.\dfrac{\left(a+b+b+c+c+a\right)^2}{a+b+b+c+c+a}=\dfrac{4}{3}\left(a+b+c\right)=\dfrac{4}{3}\)
\(P\ge3\sqrt[3]{\dfrac{abc\left(a^2+1\right)^2\left(b^2+1\right)^2\left(c^2+1\right)^2}{a^2b^2c^2\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}}=3\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{abc}}\)
\(P\ge3\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{\left(\dfrac{a+b+c}{3}\right)^3}}=9\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{\left(a+b+c\right)^3}}\ge9\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{2\left(a+b+c\right)^2}}\)
Theo nguyên lý Dirichlet, trong 3 số \(a^2;b^2;c^2\) luôn có ít nhất 2 số cùng phía so với \(\dfrac{4}{9}\)
Không mất tính tổng quát, giả sử đó là \(a^2;b^2\)
\(\Rightarrow\left(a^2-\dfrac{4}{9}\right)\left(b^2-\dfrac{4}{9}\right)\ge0\)
\(\Leftrightarrow a^2b^2+\dfrac{16}{81}\ge\dfrac{4}{9}a^2+\dfrac{4}{9}b^2\)
\(\Rightarrow a^2b^2+a^2+b^2+1\ge\dfrac{13}{9}a^2+\dfrac{13}{9}b^2+\dfrac{65}{81}\)
\(\Rightarrow\left(a^2+1\right)\left(b^2+1\right)\ge\dfrac{13}{9}\left(a^2+b^2+\dfrac{5}{9}\right)\)
\(\Rightarrow\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\dfrac{13}{9}\left(a^2+b^2+\dfrac{5}{9}\right)\left(c^2+1\right)\)
\(=\dfrac{13}{9}\left(a^2+b^2+\dfrac{4}{9}+\dfrac{1}{9}\right)\left(\dfrac{4}{9}+\dfrac{4}{9}+c^2+\dfrac{1}{9}\right)\)
\(\ge\dfrac{13}{9}\left(\dfrac{2}{3}a+\dfrac{2}{3}b+\dfrac{2}{3}c+\dfrac{1}{9}\right)^2\)
\(\Rightarrow P\ge9\sqrt[3]{\dfrac{\dfrac{13}{9}\left(\dfrac{2}{3}\left(a+b+c\right)+\dfrac{1}{9}\right)^2}{2\left(a+b+c\right)^2}}=9\sqrt[3]{\dfrac{13}{18}\left(\dfrac{2}{3}+\dfrac{1}{9\left(a+b+c\right)}\right)^2}\)
\(P\ge9\sqrt[3]{\dfrac{13}{18}\left(\dfrac{2}{3}+\dfrac{1}{9.2}\right)^2}=\dfrac{13}{2}\)
\(P_{min}=\dfrac{13}{2}\) khi \(a=b=c=\dfrac{2}{3}\)
Thầy cho em hỏi cơ sở để ta nghĩ ra dòng
\(\left(a^2-\dfrac{4}{9}\right)\left(b^2-\dfrac{4}{9}\right)\ge0\) này là gì ạ?
Theo cá nhân em thấy cách giải này hay và dễ hiểu, và có lẽ cũng dựa vào điểm rơi nhưng hình như lời giải chưa tự nhiên lắm thì phải ạ. Thầy có cách nào nữa không thầy? Em cảm ơn ạ.
\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)
\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)
\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)
\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)
\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Ta có:
\(P=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(a+b+c+36abc\right)\)
\(P=\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{b}{c}+\dfrac{c}{b}+\dfrac{a}{c}+\dfrac{c}{a}+3+36\left(ab+bc+ca\right)\)
\(P=\dfrac{a^2+b^2}{ab}+\dfrac{b^2+c^2}{bc}+\dfrac{c^2+a^2}{ca}+3+36\left(ab+bc+ca\right)\)
\(P=\dfrac{\left(a+b\right)^2}{ab}+\dfrac{\left(b+c\right)^2}{bc}+\dfrac{\left(c+a\right)^2}{ca}-3+36\left(ab+bc+ca\right)\)
\(P\ge\dfrac{4\left(a+b+c\right)^2}{ab+bc+ca}-3+36\left(ab+bc+ca\right)\)
\(P\ge\dfrac{4}{ab+bc+ca}+36\left(ab+bc+ca\right)-3\ge2\sqrt{\dfrac{4.36\left(ab+bc+ca\right)}{ab+bc+ca}}-3=21\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Lời giải:
Nếu bạn học dồn biến- thừa trừ rồi thì có thể làm như sau:
$P=\frac{ab+bc+ac}{abc}(1+36abc)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+36(ab+bc+ac)=f(a,b,c)$
Giả sử $c=\max(a,b,c)$. Ta sẽ chứng minh $f(a,b,c)\geq f(\frac{a+b}{2}, \frac{a+b}{2}, c)$
Thật vậy:
\(f(a,b,c)- f(\frac{a+b}{2}, \frac{a+b}{2}, c)=\frac{(a+b)^2-4ab}{ab(a+b)}+36.\frac{4ab-(a+b)^2}{4}\)
\(=\frac{(a-b)^2}{ab(a+b)}-9(a-b)^2=(a-b)^2(\frac{1}{ab(a+b)}-9)\)
Vì $c=\max (a,b,c)$ mà $a+b+c=1\Rightarrow a+b\leq \frac{2}{3}$
$\Rightarrow ab\leq \frac{1}{4}(a+b)^2\leq \frac{1}{9}$
$\Rightarrow \frac{1}{ab(a+b)}\geq \frac{27}{2}$
$\Rightarrow \frac{1}{ab(a+b)}-9>0$
Do đó: $f(a,b,c)\geq f(\frac{a+b}{2}, \frac{a+b}{2}, c)$
Mà:
$f(\frac{a+b}{2}, \frac{a+b}{2}, c)-21=\frac{4}{a+b}+\frac{1}{c}+36[\frac{(a+b)^2}{4}+c(a+b)]-21$
$=\frac{4}{1-c}+\frac{1}{c}+9(1-c)^2+36c(1-c)-21$
$=\frac{3c+1}{c(1-c)}+9(1-c)^2+36c(1-c)-21$
$=(3c-1)^2.\frac{3c^2-3c+1}{c(1-c)}\geq 0$ với mọi $1>c\geq \frac{1}{3}$
Do đó $f(\frac{a+b}{2}, \frac{a+b}{2}, c)\geq 21$
$\Rightarrow f(a,b,c)\geq 21$
Hay $P_{\min}=21$
\(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=1\end{matrix}\right.\) \(\Leftrightarrow0< a,b,c< 1\)
\(B=\dfrac{\left(1+a\right)\left(1+b\right)\left(1+c\right)}{\left(1-a\right)\left(1-b\right)\left(1-c\right)}=\dfrac{\left[\left(a+b\right)+\left(a+c\right)\right]\left[\left(a+b\right)+\left(b+c\right)\right]\left[\left(c+a\right)+\left(c+b\right)\right]}{\left(a+b+c-a\right)\left(a+b+c-b\right)\left(a+b+c-c\right)}\)\(\left\{{}\begin{matrix}a+b=x\\b+c=y\\c+a=z\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=2\\B=\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}\end{matrix}\right.\)
\(B>0;B^2=\dfrac{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}{\left(xyz\right)^2}=\dfrac{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}{\left(xyz\right)^2}=\dfrac{\left(x+y\right)^2}{xy}.\dfrac{\left(y+z\right)^2}{yz}.\dfrac{\left(z+x\right)^2}{zx}\)\(\left\{{}\begin{matrix}\left(x+y\right)^2\ge4xy\\\left(y+z\right)^2\ge4yz\\\left(z+x\right)^2\ge4zx\end{matrix}\right.\) \(\Leftrightarrow B^2\ge64;B\ge8\) khi x=y=z;a=b=c=1/3