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\(\text{a+b+c = 1}\Rightarrow a=1-b-c\Rightarrow a+bc=1-b-c+bc=\left(b-1\right)\left(c-1\right)\)
tương tự \(b+ca=\left(a-1\right)\left(c-1\right);c+ab=\left(a-1\right)\left(b-1\right)\)
đặt a-1=x ; b-1=y ; c-1=z , ta có
\(P=\sqrt{\frac{yzzx}{xy}}+\sqrt{\frac{xzxy}{yz}}+\sqrt{\frac{xyyz}{xz}}=\sqrt{z^2}+\sqrt{x^2}+\sqrt{y^2}=x+y+z=1\)
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\(a+bc=a\left(a+b+c\right)+bc=\left(a+b\right)\left(a+c\right)\)
Tương tự: \(b+ca=\left(a+b\right)\left(b+c\right)\) ; \(c+ab=\left(a+c\right)\left(b+c\right)\)
\(\Rightarrow P=a+b+b+c+c+a=2\left(a+b+c\right)=2\)
Ta có : \(\left\{{}\begin{matrix}a+bc=a\left(a+b+c\right)+bc=\left(a+b\right)\left(a+c\right)\\b+ca=b\left(a+b+c\right)+ca=\left(b+c\right)\left(a+b\right)\\c+ab=c\left(a+b+c\right)+ab=\left(a+c\right)\left(b+c\right)\end{matrix}\right.\)
Từ đó ta có :
\(P=\Sigma\sqrt{\frac{\left(a+b\right)\left(a+c\right)\left(b+c\right)\left(a+b\right)}{\left(a+c\right)\left(b+c\right)}}\)
\(P=\Sigma\sqrt{\left(a+b\right)^2}\)
\(P=\Sigma\left(a+b\right)\)
\(P=2\left(a+b+c\right)\)
\(P=2\)
ko cả biết BĐT AM-GM với C-S là gì còn hỏi bài này rảnh háng
Đề sai rồi. Nếu như là a, b, c dương thì giá trị nhỏ nhất của nó phải là 9 mới đúng. Còn để có GTNN như trên thì điều kiện là a, b, c không âm nhé. Mà bỏ đi e thi cái gì mà phải giải câu cỡ này. Cậu này mạnh lắm đấy không phải dạng thường đâu.
\(VT=\frac{ab+bc+ca}{ab}+\frac{ab+bc+ca}{bc}+\frac{ab+bc+ca}{ca}\)
\(=3+\frac{c\left(a+b\right)}{ab}+\frac{a\left(b+c\right)}{bc}+\frac{b\left(c+a\right)}{ca}\)(1)
Theo BĐT AM-GM: \(\frac{1}{2}\left[\frac{c\left(a+b\right)}{ab}+\frac{a\left(b+c\right)}{bc}\right]\ge\sqrt{\frac{\left(a+b\right)\left(b+c\right)}{b^2}}\)
Tương tự: \(\frac{1}{2}\left[\frac{a\left(b+c\right)}{bc}+\frac{b\left(c+a\right)}{ca}\right]\ge\sqrt{\frac{\left(a+c\right)\left(b+c\right)}{c^2}}\)
\(\frac{1}{2}\left[\frac{c\left(a+b\right)}{ab}+\frac{b\left(c+a\right)}{ca}\right]\ge\sqrt{\frac{\left(a+c\right)\left(a+b\right)}{a^2}}\)
Cộng theo vế 3 BĐT trên rồi thay vào 1 ta sẽ thu được đpcm.
Áp dụng bất đẳng thức Cauchy-Schwarz ta có:
\(\sqrt{\left(a+b\right)\left(a+c\right)}\ge\sqrt{a}.\sqrt{a}+\sqrt{b}.\sqrt{c}\)
\(\Leftrightarrow\sqrt{\left(a+b\right)\left(a+c\right)}\ge a+\sqrt{bc}\)
Do đó \(\sqrt{\frac{bc}{\left(c+a\right)\left(a+b\right)}}=\frac{\sqrt{bc\left(c+a\right)\left(a+b\right)}}{\left(c+a\right)\left(a+b\right)}\ge\sqrt{abc}\frac{\sqrt{a}}{\left(c+a\right)\left(c+b\right)}+\frac{bc}{\left(c+a\right)\left(c+b\right)}\left(1\right)\)
Chứng minh tương tự ta được:
\(\hept{\begin{cases}\sqrt{\frac{bc}{\left(c+b\right)\left(a+b\right)}}=\frac{\sqrt{bc\left(c+b\right)\left(a+b\right)}}{\left(c+b\right)\left(a+b\right)}\ge\sqrt{abc}\frac{\sqrt{b}}{\left(c+b\right)\left(a+b\right)}+\frac{ac}{\left(c+b\right)\left(a+b\right)}\left(2\right)\\\sqrt{\frac{ca}{\left(c+a\right)\left(a+b\right)}}=\frac{\sqrt{ca\left(c+a\right)\left(a+b\right)}}{\left(c+a\right)\left(a+b\right)}\ge\sqrt{abc}\frac{\sqrt{c}}{\left(c+a\right)\left(a+b\right)}+\frac{ab}{\left(a+c\right)\left(a+b\right)}\left(3\right)\end{cases}}\)
\(\Rightarrow\sqrt{\frac{bc}{\left(c+a\right)\left(a+b\right)}}+\sqrt{\frac{ca}{\left(c+b\right)\left(a+b\right)}}+\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}\ge\)
\(\sqrt{abc}\left(\frac{\sqrt{a}}{\left(a+c\right)\left(a+b\right)}+\frac{\sqrt{b}}{\left(c+b\right)\left(a+b\right)}+\frac{\sqrt{c}}{\left(c+b\right)\left(a+c\right)}\right)+\)\(\frac{bc}{\left(a+c\right)\left(a+b\right)}+\frac{ac}{\left(c+b\right)\left(a+b\right)}+\frac{ab}{\left(c+b\right)\left(a+c\right)}\left(4\right)\)
Ta lại có: \(\frac{bc}{\left(a+c\right)\left(a+b\right)}+\frac{ac}{\left(c+b\right)\left(a+b\right)}+\frac{ab}{\left(c+b\right)\left(a+c\right)}+\frac{2abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\frac{bc\left(b+c\right)+ac\left(a+c\right)+ab\left(a+b\right)+2abc}{\left(a+c\right)\left(b+c\right)\left(a+b\right)}\)
\(=\frac{bc\left(a+b+c\right)+ca\left(a+b+c\right)+ab\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\frac{c\left(a+b+c\right)\left(b+a\right)+ab\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\frac{\left(a+b\right)\left[c\left(a+c\right)+b\left(a+c\right)\right]}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\frac{\left(a+b\right)\left(c+b\right)\left(a+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=1\)
\(\left(4\right)\Leftrightarrow\sqrt{\frac{bc}{\left(c+a\right)\left(a+b\right)}}+\sqrt{\frac{ca}{\left(c+b\right)\left(a+b\right)}}+\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}\)\(\ge\sqrt{abc}\left(\frac{\sqrt{a}}{\left(c+a\right)\left(a+b\right)}+\frac{\sqrt{b}}{\left(c+b\right)\left(a+b\right)}+\frac{\sqrt{c}}{\left(c+b\right)\left(a+c\right)}\right)+1-\frac{2abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Do đó ta cần chứng minh \(\sqrt{abc}\left(\frac{\sqrt{a}}{\left(c+a\right)\left(a+b\right)}+\frac{\sqrt{b}}{\left(c+b\right)\left(a+b\right)}+\frac{\sqrt{c}}{\left(c+b\right)\left(a+c\right)}\right)+1-\frac{2abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)\(\ge1+\frac{4abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Điều này tương đương với \(\sqrt{a}\left(b+c\right)+\sqrt{b}\left(a+c\right)+\sqrt{c}\left(a+b\right)\ge6\sqrt{abc}\left(5\right)\)
Theo bất đẳng thức AM-GM thì (5) luôn đúng
Dấu "=" xảy ra khi (1);(2);(3) và (5) xảy ra dấu "=". điều này tương đương với a=b=c
Vậy ta có điều phải chứng minh
=))
\(\sqrt{\frac{\left(a+bc\right)\left(b+ac\right)}{c+ab}}=\sqrt{\frac{\left(a^2+ab+ac+bc\right)\left(b^2+bc+ba+ac\right)}{c^2+ca+cb+ab}}=\sqrt{\frac{\left(a+b\right)\left(a+c\right)\left(b+a\right)\left(b+c\right)}{\left(c+a\right)\left(c+b\right)}}=a+b\left(a,b,c>0;a+b+c=1\right)\)
Bạn làm tương tự nha
\(\Rightarrow P=a+b+c+a+b+c=2\left(a+b+c\right)=2\)
Do a + b + c = 1 nên \(\frac{\sqrt{\left(a+bc\right)\left(b+ca\right)}}{\sqrt{c+ab}}=\frac{\sqrt{\left[a\left(a+b+c\right)+bc\right]\left[b\left(a+b+c\right)+ca\right]}}{\sqrt{c\left(a+b+c\right)+ab}}\)
\(=\frac{\sqrt{\left(a^2+ab+ac+bc\right)\left(ab+b^2+bc+ac\right)}}{\sqrt{ac+bc+c^2+ab}}=\frac{\sqrt{\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)}}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)
\(=\sqrt{\left(a+b\right)^2}=a+b\) (1)
Tương tự \(\hept{\begin{cases}\frac{\sqrt{\left(b+ca\right)\left(c+ab\right)}}{\sqrt{a+bc}}=b+c\text{ }\left(2\right)\\\frac{\sqrt{\left(c+ab\right)\left(a+bc\right)}}{\sqrt{b+ac}}=a+c\text{ }\left(3\right)\end{cases}}\)
Cộng vế với vế của (1)(2)(3) lại ta được :
\(\frac{\sqrt{\left(a+bc\right)\left(b+ca\right)}}{\sqrt{c+ab}}+\frac{\sqrt{\left(b+ca\right)\left(c+ab\right)}}{\sqrt{a+bc}}+\frac{\sqrt{\left(c+ab\right)\left(a+bc\right)}}{\sqrt{b+ac}}=2\left(a+b+c\right)=2\)