TA CÓ:
\(Q=\frac{x\left(\sqrt{x+zy}-x\right)}{x+yz-x^2}+\frac{y\left(\sqrt{y+zx}-y\right)}{y+zx-y^2}+\frac{z\left(\sqrt{xy+z}-z\right)}{z+xy-z^2}\)

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

\(=\frac{x\left(\sqrt{\left(x+y\right)\left(z+x\right)}-x\right)}{xy+yz+zx}+\frac{y\left(\sqrt{\left(x+y\right)\left(y+z\right)}-y\right)}{xy+yz+zx}+\frac{z\left(\sqrt{\left(y+z\right)\left(z+x\right)}-z\right)}{xy+yz+za}\)

ÁP DỤNG BĐT CÔ-SI TA ĐƯỢC:

\(Q\le\frac{x\left(\frac{x+y+z+x}{2}-x\right)}{xy+zx+yz}+\frac{y\left(\frac{x+y+z+y}{2}-y\right)}{xy+yz+zx}+\frac{z\left(\frac{x+y+z+z}{2}-z\right)}{xy+yz+zx}\)

\(=\frac{xy+zx}{2\left(xy+yz+zx\right)}+\frac{xy+yz}{2\left(xy+yz+zx\right)}+\frac{yz+zx}{2\left(xy+yz+zx\right)}=1\)

DẤU BẰNG  XẢY RA \(\Leftrightarrow x=y=z=\frac{1}{3}\)

=

1/3

hok tốt

Ta có: \(P=\frac{\sqrt{yz}}{x+2\sqrt{yz}}+\frac{\sqrt{zx}}{y+2\sqrt{zx}}+\frac{\sqrt{xy}}{z+2\sqrt{xy}}=\frac{1}{\frac{x}{\sqrt{yz}}+2}+\frac{1}{\frac{y}{\sqrt{zx}}+2}+\frac{1}{\frac{z}{\sqrt{xy}}+2}\)

Đặt \(\frac{x}{\sqrt{yz}}=c,\frac{y}{\sqrt{zx}}=t;\frac{z}{\sqrt{xy}}=k\left(c,t,k>0\right)\)thì ctk = 1

Ta cần tìm giá trị lớn nhất của \(P=\frac{1}{c+2}+\frac{1}{t+2}+\frac{1}{k+2}\)với ctk = 1

Dự đoán MaxP = 1 khi c = t = k = 1

Thật vậy: \(P=\frac{kt+2k+2t+4+ct+2c+2t+4+ck+2c+2k+4}{\left(c+2\right)\left(t+2\right)\left(k+2\right)}=\frac{\left(kt+tc+ck\right)+4\left(c+t+k\right)+12}{ctk+2\left(kt+tc+ck\right)+4\left(c+t+k\right)+8}\le\frac{\left(kt+tc+ck\right)+4\left(c+t+k\right)+12}{1+\left(kt+tc+ck\right)+3\sqrt[3]{\left(ctk\right)^2}+4\left(c+t+k\right)+8}=1\)Đẳng thức xảy ra khi x = y = z

Ta có: \(\frac{\sqrt{yz}}{x+2\sqrt{yz}}=\frac{1}{2}\left(1-\frac{x}{x+2\sqrt{yz}}\right)\le\frac{1}{2}\left(1-\frac{x}{x+y+z}\right)=\frac{1}{2}\left(\frac{y+z}{x+y+z}\right)\)(bđt cosi) (1)

CMTT: \(\frac{\sqrt{xz}}{y+2\sqrt{xz}}\le\frac{1}{2}\left(\frac{x+z}{x+y+z}\right)\)(2)

\(\frac{\sqrt{xy}}{z+2\sqrt{xy}}\le\frac{1}{2}\left(\frac{x+y}{x+y+z}\right)\)(3)

Từ (1), (2) và (3) cộng vế theo vế ta có:

\(\frac{\sqrt{yz}}{x+2\sqrt{yz}}+\frac{\sqrt{xz}}{y+2\sqrt{xz}}+\frac{\sqrt{xy}}{z+2\sqrt{xy}}\le\frac{1}{2}\left(\frac{y+z}{x+y+z}\right)+\frac{1}{2}\left(\frac{x+z}{x+y+z}\right)+\frac{1}{2}\left(\frac{x+y}{x+y+z}\right)\)

=> P \(\le\frac{1}{2}\left(\frac{y+z+x+z+x+y}{x+y+z}\right)=\frac{1}{2}\cdot\frac{2\left(x+y+z\right)}{x+y+z}=1\)

Dấu "=" xảy ra <=> x = y = z

Vậy MaxP = 1 <=> x = y = z

một bài khá hay :)

Ta có \(\frac{\sqrt{yz}}{x+2\sqrt{yz}}=1-\frac{x}{x+2\sqrt{yz}}\le1-\frac{x}{x+y+z}\left(1\right)\)

 Tương tự \(\frac{\sqrt{xz}}{y+2\sqrt{xz}}=1-\frac{y}{y+2\sqrt{xz}}\le1-\frac{y}{x+y+z}\left(2\right)\)

\(\frac{\sqrt{xy}}{z+2\sqrt{xy}}=1-\frac{z}{z+2\sqrt{xy}}\le1-\frac{z}{x+y+z}\left(3\right)\)

Cộng (1);(2);(3)

\(2P\le3-\frac{x+y+z}{x+y+z}=2\Rightarrow P\le1\)

Vậy \(minP=1\)Khi và chỉ khi \(x=y=z\)

Câu hỏi của phan tuấn anh - Toán lớp 9 - Học toán với OnlineMath cái này y hệt, tham khảo đi nếu vẫn chưa làm dc thì nhắn cho mk

Theo giả thiết \(\sqrt{\frac{yz}{x}}+\sqrt{\frac{xz}{y}}+\sqrt{\frac{xy}{z}}=3\)

\(\Rightarrow\frac{yz}{x}+\frac{xz}{y}+\frac{xy}{z}+2x+2y+2z=9\)

Mặt khác , ta có BĐT phụ : \(\frac{yz}{x}+\frac{xz}{y}+\frac{xy}{z}\ge x+y+z\)

\(\Rightarrow9\ge3\left(x+y+z\right)\)

\(\Leftrightarrow x+y+z\le3\)

Áp dụng BĐT Cauchy Shwarz \(\Rightarrow\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2\le3\left(x+y+z\right)\le9\)

\(\Rightarrow\sqrt{x}+\sqrt{y}+\sqrt{z}\le3\)

Ta có : \(P=\sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{2016}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

\(=\sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{9}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\frac{2007}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

\(\ge2.\sqrt{9}+\frac{2007}{3}=675\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)

Chúc bạn học tốt !!!

Từ \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1\)

\(\Rightarrow\)\(x+y+z=xyz\)

Ta có : \(\sqrt{yz\left(1+x^2\right)}=\sqrt{yz+x^2yz}=\sqrt{yz+x\left(x+y+z\right)}=\sqrt{\left(x+y\right)\left(x+z\right)}\)

Tương tự : \(\sqrt{xy\left(1+z^2\right)}=\sqrt{\left(z+y\right)\left(z+x\right)}\); \(\sqrt{zx\left(1+y^2\right)}=\sqrt{\left(y+z\right)\left(y+x\right)}\)

Nên \(Q=\frac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}+\frac{y}{\sqrt{\left(y+z\right)\left(y+x\right)}}+\frac{z}{\sqrt{\left(z+x\right)\left(z+y\right)}}\)

         \(Q=\sqrt{\frac{x}{x+y}.\frac{x}{x+z}}+\sqrt{\frac{y}{x+y}.\frac{y}{y+z}}+\sqrt{\frac{z}{x+z}.\frac{z}{y+z}}\)

Áp dụng BĐT \(\sqrt{A.B}\le\frac{A+B}{2}\left(A,B>0\right)\)

Dấu "=" xảy ra khi A = B :

Ta được :

\(Q\le\frac{1}{2}\left(\frac{x}{x+y}+\frac{x}{x+z}+\frac{y}{y+x}+\frac{y}{y+z}+\frac{z}{z+x}+\frac{z}{z+y}\right)=\frac{3}{2}\)

Vậy GTLN của \(Q=\frac{3}{2}\)khi \(x=y=z=\sqrt{3}\)

\(3=x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\)

\(\Rightarrow xyz\le1\)

\(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\le\frac{x^2+1+1}{3}+\frac{y^2+1+1}{3}+\frac{z^2+1+1}{3}=3\)

Ta co:

\(A=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}=\frac{x\sqrt[3]{x}}{\sqrt[3]{xyz}}+\frac{y\sqrt[3]{y}}{\sqrt[3]{xyz}}+\frac{z\sqrt[3]{z}}{\sqrt[3]{xyz}}\)

\(\ge x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\)

\(\Rightarrow3A\ge3\left(x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\right)\ge\left(x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\right)\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\right)\)

\(\ge\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)

\(\Rightarrow A\ge xy+yz+zx\)

Áp dụng BĐT Cauchy - Schwarz, ta có: \(3\left(x^2+y^2+z^2\right)=\left(1^2+1^2+1^2\right)\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)

\(\Rightarrow x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=3=x^2+y^2+z^2\)(Do \(x^2+y^2+z^2=3\))

Ta có: \(\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{zx}}+\frac{z}{\sqrt[3]{xy}}=\frac{x}{\sqrt[3]{yz.1}}+\frac{y}{\sqrt[3]{zx.1}}+\frac{z}{\sqrt[3]{xy.1}}\)

\(\ge\frac{x}{\frac{y+z+1}{3}}+\frac{y}{\frac{z+x+1}{3}}+\frac{z}{\frac{x+y+1}{3}}\)\(=\frac{3x}{y+z+1}+\frac{3y}{z+x+1}+\frac{3z}{x+y+1}\)

\(=\frac{3x^2}{xy+zx+x}+\frac{3y^2}{yz+xy+y}+\frac{3z^2}{zx+yz+z}\)\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+\left(x+y+z\right)}\)(Theo BĐT Cauchy - Schwarz dạng Engle)

\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+x^2+y^2+z^2}=\frac{3\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=3=x^2+y^2+z^2\)

\(\ge xy+yz+zx\)

Đẳng thức xảy ra khi x = y = z = 1

\(\sqrt[3]{yz\cdot1}\le\frac{y+z+1}{3};\sqrt[3]{xz\cdot1}\le\frac{x+z+1}{3};\sqrt[3]{yx\cdot1}\le\frac{y+x+1}{3}\)

Nên \(A=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}\ge3\left(\frac{x}{y+z+1}+\frac{y}{x+z+1}+\frac{z}{y+x+1}\right)\)\(=3\left(\frac{x^2}{xy+yz+x}+\frac{y^2}{xy+yz+y}+\frac{z^2}{yz+xz+z}\right)=B\)

\(B\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+x+y+z}\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+x^2+y^2+z^2}=\frac{3\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=3\ge xy+yz+zx\)

do \(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)=9\Rightarrow x+y+z\le3=x^2+y^2+z^2;xy+yz+zx\le x^2+y^2+z^2=3\)

ta có 3x + yz = x² + xy + yz + zx = (x+y)(x+z)

do đó:

\(\frac{x}{x+\sqrt{3x+yz}}=\frac{x\left(\sqrt{x^2+xy+yz+zx}-x\right)}{\left(\sqrt{x^2+xy+yz+zx}+x\right)\left(\sqrt{x^2+xy+yz+zx}-x\right)}\)

= \(\frac{x\left(\sqrt{\left(x+y\right)\left(x+z\right)}-x\right)}{xy+yz+zx}\le\frac{x\left(\frac{x+y+x+z}{2}-x\right)}{xy+yz+zx}\)\(\le\frac{x\left(y+z\right)}{2\left(xy+yz+zx\right)}\)

tương tự với 2 số hạng còn lại nên ta được: P\(\le\)1. đpcm

hi minh ket ban nhe

m.imgur.com/a/ls9dmpn

Cậu chịu khó đánh máy nhé ! Tớ dùng đt nên nhác phải đánh text lắm :(((

Cách mình ngắn hơn trên khá nhìu nha !!!!

Lời giải:

Do $x+y+z=1$ nên:

$x+yz=x(x+y+z)+yz=(x+y)(x+z)=(x+y)(z+x)\geq (\sqrt{xz}+\sqrt{xy})^2$ theo BĐT Bunhiacopxky

$\Rightarrow \sqrt{x+yz}\geq \sqrt{xz}+\sqrt{xy}$

$\Rightarrow \frac{x}{x+\sqrt{x+yz}}\leq \frac{x}{x+\sqrt{xz}+\sqrt{xy}}=\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}$

Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế suy ra:

$Q\leq \frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1$

Vậy $Q_{\max}=1$ khi $x=y=z=\frac{1}{3}$