Áp dụng BĐT Cô - si cho 2 số \(\frac{xy}{z};\frac{yz}{x}\)dương ta có: \(\frac{xy}{z}+\frac{yz}{x}\ge2\sqrt{\frac{xy}{z}.\frac{yz}{x}}=2\sqrt{y^2}=2y\)(1)

Tương tự. \(\frac{yz}{x}+\frac{zx}{y}\ge2\sqrt{\frac{yz}{x}.\frac{zx}{y}}=2\sqrt{z^2}=2z\) (2);

\(\frac{xy}{z}+\frac{zx}{y}\ge2\sqrt{\frac{xy}{z}.\frac{zx}{y}}=2\sqrt{x^2}=2x\)(3)

Cộng từng vế của (1)(2)(3) ta được \(2.\left(\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\right)\ge2\left(x+y+z\right)=2\Rightarrow P\ge1\)

Vậy Min P = 1 tại x= y = z = 1/3

khó quá!!!!!!!!!!!

Theo em bài này chỉ có min thôi nhé!

Rất tự nhiên để khử căn thức thì ta đặt \(\left(\sqrt{x};\sqrt{y};\sqrt{z}\right)=\left(a;b;c\right)\ge0\)

Khi đó \(M=\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\) với abc = \(\sqrt{xyz}=1\) và a,b,c > 0

Dễ thấy \(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\)

(chuyển vế qua dùng hằng đẳng thức là xong liền hà)

Do đó \(2M=\frac{a^3+b^3}{a^2+ab+b^2}+\frac{b^3+c^3}{b^2+bc+c^2}+\frac{c^3+a^3}{c^2+ca+a^2}\)

Đến đây thì chứng minh \(\frac{a^3+b^3}{a^2+ab+b^2}\ge\frac{1}{3}\left(a+b\right)\Leftrightarrow\frac{2}{3}\left(a-b\right)^2\left(a+b\right)\ge0\)(đúng)

Áp dụng vào ta thu được: \(2M\ge\frac{2}{3}\left(a+b+c\right)\Rightarrow M\ge\frac{1}{3}\left(a+b+c\right)\ge\sqrt[3]{abc}=1\)

Vậy...

P/s: Ko chắc nha!

dit me may 

bạn bui thai hoc sao lại cmt linh tinh vậy :)) bạn ko có học thức à :> mà ý bạn cmt như vậy là sao hả ? 

Cách 1:

Ta có \(A=xy+yz+2zx\)

\(\Rightarrow A+1=x^2+y^2+z^2+xy+yz+2zx\)

                    \(=\left(x+z+\frac{y}{2}\right)^2+\frac{3}{4}y^2\ge0\)

\(\Rightarrow A\ge-1\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}y=0\\x=-z\end{cases}}\)

Ta có : \(\left(x+y+z\right)^2\ge0\)

\(\Rightarrow xy+yz+zx\ge\frac{-\left(x^2+y^2+z^2\right)}{2}=-\frac{1}{2}\)

Lại có : \(\left(x+z\right)^2\ge0\Rightarrow xz\ge\frac{-\left(x^2+z^2\right)}{2}=\frac{y^2-1}{2}\ge-\frac{1}{2}\)

Khi đó : \(xy+yz+2zx\ge-1\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}y=o\\x^2=z^2=\frac{1}{2}\end{cases}}\)

Cách 2

Ta có : \(\left(x+y+z\right)^2\ge0\)

\(\Leftrightarrow xy+yz+zx\ge\frac{-\left(x^2+y^2+z^2\right)}{2}=\frac{-1}{2}\)

Mặt khác: \(\left(x+z\right)^2\ge0\Rightarrow xz\ge\frac{-\left(x^2+z^2\right)}{2}=\frac{y^2-1}{2}\ge\frac{-1}{2}\)

\(\Rightarrow xy+yz+2xz\ge-1\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}y=0\\x=-z\end{cases}}\)

Lời giải:

Đặt $\left(\frac{xy}{z};\frac{yz}{x};\frac{xz}{y}\right)=(a,b,c)$

$\Rightarrow ⎧⎪⎨⎪⎩\begin{array}{c}{}^{}=ab\\ {}^{}=ac\\ {}^{}=bc\end{array}$

Bài toán trở thành: Cho $a,b,c>0$ thỏa mãn $ab+bc+ac=1$

Tìm min $S=a+b+c$

Theo hệ quả quen thuộc của BĐT Cauchy: $(a+b+c{}^{}\ge 3(ab+bc+ac)$

$\Rightarrow S=\sqrt{(a+b+c{}^{}}\ge \sqrt{3(ab+bc+ac)}=\sqrt{3}$

Vậy Smin=√3⇔a=b=c=13⇔x=y=z=1√3

Bạn bôi xanh câu hỏi của bạn rồi kéo thả lên chỗ tìm kiếm ; tìm 

 OK !

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P/S : Mình ghi lộn biến rồi, bạn cứ đổi (a;b;c) thành (x;y;z) đi, còn (x;y;z) thành (a;b;c).