Đề có sai không bạn?

ko làm đc thì chắc là sai thôi bạn

\(M=\frac{2x+y+z-15}{x}+\frac{x+2y+z-15}{y}+\frac{x+y+2z-15}{z}\)

\(M-3=\frac{x+y+z-15}{x}+\frac{x+y+z-15}{y}+\frac{x+y+z-15}{z}\)

\(M-3=\left(x+y+z-15\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(\Rightarrow M\ge\left(x+y+z-15\right)\cdot\frac{9}{x+y+z}+3=\frac{3}{4}\)

\("="\Leftrightarrow x=y=z=4\)

nhận ra là bài này sai đề :)))

nhận ra là bài này sai đề :)))

Bài 1

M=2x+y+z−15x+x+2y+z−15y+x+y+2z−15z

M=x+12−15x+y+12−15y+z+12−15z

M=x−3x+y−3y+z−3z

M=1−3x+1−3y+1−3z

M=3−(3x+3y+3z)

M=3−3(1x+1y+1z)

Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức

⇒1x+1y+1z≥(1+1+1)2x+y+z=9x+y+z=34

⇒3(1x+1y+1z)≥94

⇒3−3(1x+1y+1z)≤34

⇔M≤34

Vậy M max=34

Dấu " = " xảy ra khi x=y=z=4

Bai nay tim GTLN moi dung nha

Câu a :

Ta có : \(\sqrt{5+3x}-\sqrt{5-3x}=a\)

\(\Leftrightarrow\left(\sqrt{5+3x}-\sqrt{5-3x}\right)^2=a^2\)

\(\Leftrightarrow5+3x-2\sqrt{\left(5+3x\right)\left(5-3x\right)}+5-3x=a^2\)

\(\Leftrightarrow10-2\sqrt{25-9x^2}=a^2\)

\(\Leftrightarrow2\sqrt{25-9x^2}=10-a^2\)

\(\Leftrightarrow\sqrt{25-9x^2}=\dfrac{10-a^2}{2}\)

\(\Leftrightarrow25-9x^2=\dfrac{\left(a^2-10\right)^2}{2}\)

\(\Leftrightarrow9x^2=25-\dfrac{\left(a^2-10\right)^2}{2}\)

\(\Leftrightarrow3x=\sqrt{\dfrac{50-\left(a^2-10\right)^2}{2}}\)

\(\Leftrightarrow x=\dfrac{\sqrt{50-\left(a^2-10\right)^2}}{3\sqrt{2}}\)

\(P=\dfrac{3\sqrt{2}.\sqrt{10+2\sqrt{\dfrac{10-a^2}{2}}}}{\sqrt{50-\left(a^2-10\right)^2}}\)

Bạn tự rút gọn nữa nhé :))

Câu b : \(M=\dfrac{2x+y+z-15}{x}+\dfrac{x+2y+z-15}{y}+\dfrac{x+y+2z-24}{z}\)

\(=\dfrac{x-3}{x}+\dfrac{y-3}{y}+\dfrac{z-12}{z}\)

\(=3-3\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{4}{z}\right)\le3-3\left[\dfrac{\left(1+1+2\right)^2}{12}\right]=-1\)

Bài 1 :

Ta có : \(\dfrac{1}{3a^2+b^2}+\dfrac{2}{b^2+3ab}=\dfrac{1}{3a^2+b^2}+\dfrac{4}{2b^2+6ab}\)

Theo BĐT Cô - Si dưới dạng engel ta có :

\(\dfrac{1}{3a^2+b^2}+\dfrac{4}{2b^2+6ab}\ge\dfrac{\left(1+2\right)^2}{3a^2+6ab+3b^2}=\dfrac{9}{3\left(a+b\right)^2}=\dfrac{9}{3.1}=3\)

Dấu \("="\) xảy ra khi : \(a=b=\dfrac{1}{2}\)

Áp dụng bđt Cauchy-Schwarz:

\(A=\dfrac{1}{\sqrt{x\left(y+2z\right)}}+\dfrac{1}{\sqrt{y\left(z+2x\right)}}+\dfrac{1}{\sqrt{z\left(x+2y\right)}}\)

\(\ge\dfrac{\left(1+1+1\right)^2}{\sqrt{x\left(y+2z\right)}+\sqrt{y\left(z+2x\right)}+\sqrt{z\left(x+2y\right)}}\)

\(=\dfrac{9}{\sqrt{x\left(y+2z\right)}+\sqrt{y\left(z+2x\right)}+\sqrt{z\left(x+2y\right)}}\)

Áp dụng liên tiếp Bunyakovsky và AM-GM:

\(\left(\sqrt{x\left(y+2z\right)}+\sqrt{y\left(z+2x\right)}+\sqrt{z\left(x+2y\right)}\right)^2\)

\(\le\left(1^2+1^2+1^2\right)\left[x\left(y+2z\right)+y\left(z+2x\right)+z\left(x+2y\right)\right]\)

\(=3.3\left(xy+yz+xz\right)\)

Mà \(3\left(xy+yz+xz\right)\le\left(x+y+z\right)^2=3\)

\(3.3\left(xy+yz+xz\right)\le3.3=9\)

\(\Leftrightarrow\sqrt{x\left(y+2z\right)}+\sqrt{y\left(z+2x\right)}+z\sqrt{\left(x+2y\right)}\le\sqrt{9}=3\)

\(\Leftrightarrow A\ge\dfrac{9}{3}=3."="\Leftrightarrow x=y=z=\dfrac{1}{\sqrt{3}}\)

Áp dụng bất đẳng thức cauchy:

\(P=\sum\dfrac{x^2\left(y+z\right)}{y\sqrt{y}+2z\sqrt{z}}\ge\sum\dfrac{2x^2\sqrt{yz}}{y\sqrt{y}+2z\sqrt{z}}=\sum\dfrac{2\sqrt{x^3}\sqrt{xyz}}{\sqrt{y^3}+2\sqrt{z^3}}=\sum\dfrac{2\sqrt{x^3}}{\sqrt{y^3}+2\sqrt{z^3}}\)(vì xyz=1).

đặt \(\left\{{}\begin{matrix}\sqrt{x^3}=a\\\sqrt{y^3}=b\\\sqrt{z^3}=c\end{matrix}\right.\)(\(a,b,c>0\))thì giả thiết trở thành cho abc=1. tìm Min \(P=\dfrac{2a}{b+2c}+\dfrac{2b}{c+2a}+\dfrac{2c}{a+2b}\)

Áp dụng BĐT cauchy-schwarz:

\(P=2\left(\dfrac{a^2}{ab+2ac}+\dfrac{b^2}{bc+2ab}+\dfrac{c^2}{ac+2bc}\right)\ge\dfrac{2\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\ge\dfrac{2\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=2\)( AM-GM \(3\left(ab+bc+ca\right)\le\left(a+b+c\right)^2\))

Dấu = xảy ra khi a=b=c=1 hay x=y=z=1

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\(\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)(x+x+y+z)\geq (1+1+1+1)^2\)

\(\Rightarrow \frac{2}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{16}{2x+y+z}\)

Hoàn toàn tương tự:

\(\frac{1}{x}+\frac{2}{y}+\frac{1}{z}\geq \frac{16}{x+2y+z}\)

\(\frac{1}{x}+\frac{1}{y}+\frac{2}{z}\geq \frac{16}{x+y+2z}\)

Cộng theo vế các BĐT vừa thu được:

\(\Rightarrow 4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\geq 16\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)

\(\Rightarrow 16\geq 16\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)

\(\Rightarrow \frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\leq 1\)

Ta có đpcm.

Ta có :

\(\dfrac{1}{2x+y+z}=\dfrac{16}{16\left(x+x+y+z\right)}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

\(\dfrac{1}{x+2y+z}=\dfrac{16}{16\left(x+y+y+z\right)}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

\(\dfrac{1}{x+y+2z}=\dfrac{16}{16\left(x+y+z+z\right)}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}\right)\)

Cộng từng vế của BĐT ta được :

\(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{4}{x}+\dfrac{4}{y}+\dfrac{4}{z}\right)=\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=1\)

Vậy BĐT đã được chứng minh !