1. 1/x + 2/1-x = (1/x - 1) + (2/1-x - 2) + 3

= 1-x/x + (2-2(1-x))/1-x  + 3

= 1-x/x + 2x/1-x + 3    >= 2√2 + 3

Dấu "=" xảy ra khi x =√2 - 1

2. a = √z-1, b = √x-2, c = √y-3 (a,b,c >=0)

=> P = √z-1 / z + √x-2 / x + √y-3 / y 

= a/a^2+1 + b/b^2+2 + c/c^2+3

a^2+1 >= 2a              => a/a^2+1 <= 1/2

b^2+2 >= 2√2 b          => b/b^2+2 <= 1/2√2

c^2+3 >= 2√3 c            => c/c^2+3 <= 1/2√3

=> P <= 1/2 + 1/2√2 + 1/2√3

Dấu = xảy ra khi a^2 = 1, b^2 = 2, c^2 =3

<=> z-1 = 1, x-2 = 2, y-3 = 3

<=> x=4, y=6, z=2

Lời giải:

Theo hệ quả của BĐT AM-GM:

\(x^2+y^2+z^2\geq xy+yz+xz\)

\(\Leftrightarrow (x+y+z)^2\geq 3(xy+yz+xz)\Leftrightarrow xy+yz+xz\leq 3\)

Do đó:

\(P=\sum \frac{xy}{\sqrt{z^2+3}}\leq \sum \frac{xy}{\sqrt{z^2+xy+yz+xz}}\)

\(\Leftrightarrow P\leq \sum \frac{xy}{\sqrt{(z+x)(z+y)}}\) (1)

Áp dụng BĐT AM-GM:

\(\frac{2xy}{\sqrt{(z+x)(z+y)}}\leq \frac{xy}{z+x}+\frac{xy}{z+y}\)

\(\frac{2yz}{\sqrt{(y+x)(x+z)}}\leq \frac{yz}{y+x}+\frac{yz}{x+z}\)

\(\frac{2xz}{\sqrt{(x+y)(y+z)}}\leq \frac{xz}{x+y}+\frac{xz}{z+y}\)

Cộng theo vế:

\(2\sum \frac{xy}{\sqrt{(z+x)(z+y)}}\leq \frac{y(x+z)}{x+z}+\frac{x(y+z)}{y+z}+\frac{z(x+y)}{x+y}\)

\(\Leftrightarrow 2\sum \frac{xy}{\sqrt{(z+y)(z+x)}}\leq x+y+z=3\)

\(\Leftrightarrow \sum \frac{xy}{\sqrt{(z+y)(z+x)}}\leq \frac{3}{2}(2)\)

Từ \((1);(2)\Rightarrow P\leq \frac{3}{2}\Leftrightarrow P_{\max}=\frac{3}{2}\)

Dấu bằng xảy ra khi \(x=y=z=1\)

Câu hỏi của Vo Trong Duy - Toán lớp 9 - Học toán với OnlineMath

Giải :

Ta có : \(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\Leftrightarrow\dfrac{x}{10}=\dfrac{y}{15}\\\dfrac{y}{5}=\dfrac{z}{7}\Leftrightarrow\dfrac{y}{15}=\dfrac{z}{21}\end{matrix}\right.\)

\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}\)

Đặt \(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=k\Rightarrow\left\{{}\begin{matrix}x=10k\\y=15k\\z=21k\end{matrix}\right.\)

Thay x = 10k ; y = 15k và z = 21k vào x + y + z = 92 ta có :

10k + 15k + 21k = 92

⇔ k ( 10+15+21) = 92

\(\Leftrightarrow k\cdot46=92\\ \Leftrightarrow k=2\\ \Rightarrow\left\{{}\begin{matrix}x=10\cdot2=20\\y=15\cdot2=30\\z=21\cdot2=42\end{matrix}\right.\\ Vậyx=20;y=30vàz=42\)

ta có :\(\dfrac{x}{2}=\dfrac{y}{3}=>\dfrac{x}{10}=\dfrac{y}{15}\left(1\right)\)

lại có :\(\dfrac{y}{5}=\dfrac{z}{7}=>\dfrac{y}{15}=\dfrac{z}{21}\left(2\right)\)

từ (1)và (2) => \(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{x+y+z}{10+15+21}=\dfrac{92}{46}=2\)

=>\(\left\{{}\begin{matrix}x=20\\y=30\\z=42\end{matrix}\right.\)

chúc bạn học tốt

\(x^2+4xy+y^2\ge6xy\Leftrightarrow4\left(x^2+xy+y^2\right)\ge3\left(x+y\right)^2\Leftrightarrow\sqrt{x^2+xy+y^2}\ge\dfrac{\sqrt{3}}{2}\left(x+y\right)\)

\(\Rightarrow P=\sum\sqrt{x^2+xy+y^2}\ge\dfrac{\sqrt{3}}{2}\sum\left(x+y\right)=\sqrt{3}\left(x+y+z\right)=3\)

\(\Rightarrow P_{min}=3\) khi \(x=y=z=\dfrac{\sqrt{3}}{3}\)

Phá đi

Xem XXX ko mấy man

Hi anh trai, nhớ em là ai chứ :))

Áp dụng BĐT AM - GM: \(x+y+z\ge3\sqrt[3]{xyz}=3\)

\(P=\Sigma\dfrac{1}{\left(3x+1\right)\left(y+z\right)+x}\) \(=\Sigma\dfrac{1}{3x\left(y+z\right)+x+y+z}\)

\(\Rightarrow P\le\Sigma\dfrac{1}{3x\left(y+z\right)+3}\)

\(\Leftrightarrow3P\le\Sigma\dfrac{1}{x\left(y+z\right)+1}\)

Chia cả hai vế cho \(xyz=1\)

\(\Leftrightarrow3P\le\Sigma\dfrac{1}{\dfrac{1}{y}+\dfrac{1}{z}+1}\)

Đặt \(a=\sqrt[3]{\dfrac{1}{x^3}},b=\sqrt[3]{\dfrac{1}{y^3}},c=\sqrt[3]{\dfrac{1}{z^3}}\)

\(\Rightarrow a.b.c=1\)

\(\Rightarrow3P\le\Sigma\dfrac{1}{a^3+b^3+1}\)

Mặt khác: \(\left(a-b\right)^2\ge0\)

\(\Leftrightarrow a^2-ab+b^2\ge ab\)

Nhân cả hai vế cho \(a+b\)

\(\Leftrightarrow a^3+b^3\ge ab\left(a+b\right)\)

\(\Leftrightarrow a^3+b^3+1\ge ab\left(a+b\right)+1=ab\left(a+b\right)+abc\)

\(\Leftrightarrow a^3+b^3+1\ge ab\left(a+b+c\right)\)

\(\Leftrightarrow3P\le\Sigma\dfrac{1}{ab\left(a+b+c\right)}=1\)

\(\Leftrightarrow P\le\dfrac{1}{3}\)

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

Áp dụng t/c dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=\dfrac{2y}{4a+2b-2c}=\dfrac{x+2y+z}{a+2b+c+4a+2b-2c+4a-4b+c}=\dfrac{x+2y+z}{9a}\)

\(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=\dfrac{2x}{2a+4b+2c}=\dfrac{2x+y-z}{2a+4b+2x+2a+b-c-4a+4b-c}=\dfrac{2x+y+z}{9b}\)

\(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=\dfrac{4x}{4a+8b+4c}=\dfrac{4y}{8a+4b-4c}=\dfrac{4x-4y+z}{4a+8b+4c-8a-4b+4c+4a-4b+c}=\dfrac{4x-4y-z}{9c}\)

=> \(\dfrac{x+2y+z}{9a}=\dfrac{2x+y+z}{9b}=\dfrac{4x-4y+z}{9c}\)

=> \(\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y+z}=\dfrac{c}{4x-4y+z}\)