Cho x,y,z >0 thỏa mãn \(x+y+z=3\)
Tìm min \(Q=\sqrt[3]{\frac{x^5+y^5}{x^2+y^2}}+\sqrt[3]{\frac{y^5+z^5}{y^2+z^2}}+\sqrt[3]{\frac{z^5+x^5}{z^2+x^2}}\)
a,GPT \(\sqrt{x^3+12}-3x=\sqrt{x^2+5}-5\)
b,Cho x,y,z>0 thỏa mãn \(\sqrt{x}+\sqrt{y}+\sqrt{z}=3\)TÌM GTNN \(P=\frac{\sqrt{x}+1}{y+1}+\frac{\sqrt{y}+1}{z+1}+\frac{\sqrt{z}+1}{x+1}\)
b, Ta có
\(\frac{\sqrt{x}+1}{y+1}=\frac{\left(\sqrt{x}+1\right)\left(y+1\right)-y-y\sqrt{x}}{y+1}=\sqrt{x}+1-\frac{y\left(\sqrt{x}+1\right)}{y+1}\)
Mà \(y+1\ge2\sqrt{y}\)
=> \(\frac{\sqrt{x}+1}{y+1}\ge\sqrt{x}+1-\frac{1}{2}\sqrt{y}\left(\sqrt{x}+1\right)\)
Khi đó
\(P\ge\frac{1}{2}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)+3-\frac{1}{2}\left(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\right)\)
Mà \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\le\frac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2}{3}=3\)
=> \(P\ge\frac{1}{2}.3+3-\frac{3}{2}=3\)
Vậy MinP=3 khi x=y=z=1
Tìm Min của: \(\sqrt{\frac{x^3}{x^3+8y^3}}+\sqrt{\frac{4y^3}{y^3+\left(x+y\right)^3}}\) với x,y>0
Cho xy+yz+zx=5 x,y,z>0
Tìm Min của A= \(\frac{3x+3y+3z}{\sqrt{6\left(x^2+5\right)}+\sqrt{6\left(y^2+5\right)}+\sqrt{z^2+5}}\)
1. gpt : \(\frac{2x+1}{\sqrt{x^2+2}}+\left(x+1\right)\sqrt{1+\frac{2x+1}{x^2+2}}+x=0\)
2. \(\left\{{}\begin{matrix}x,y,z>0\\x+y+z\le\frac{3}{2}\end{matrix}\right.\) Tìm min \(Q=\frac{x}{y^2z}+\frac{y}{z^2x}+\frac{z}{x^2y}+\frac{x^5}{y}+\frac{y^5}{z}+\frac{z^5}{x}\)
a/ \(\frac{2x+1}{\sqrt{x^2+2}}+\left(x+1\right)\left(\sqrt{1+\frac{2x+1}{x^2+2}}-1\right)+2x+1=0\)
\(\Leftrightarrow\frac{2x+1}{\sqrt{x^2+2}}+\frac{\left(x+1\right)\left(2x+1\right)}{\sqrt{1+\frac{2x+1}{x^2+2}}+1}+2x+1=0\)
\(\Leftrightarrow\left(2x+1\right)\left(\frac{1}{\sqrt{x^2+2}}+\frac{x+1}{\sqrt{1+\frac{2x+1}{x^2+2}}+1}+1\right)=0\)
\(\Rightarrow x=-\frac{1}{2}\)
b/ \(Q\ge\frac{\left(x+y+z\right)^2}{xyz\left(x+y+z\right)}+\frac{\left(x^3+y^3+z^3\right)^2}{xy+yz+zx}\ge\frac{x+y+z}{xyz}+\frac{\left(x^2+y^2+z^2\right)^3}{\left(x+y+z\right)^2}\)
\(Q\ge\frac{27\left(x+y+z\right)}{\left(x+y+z\right)^3}+\frac{\left(x+y+z\right)^6}{27\left(x+y+z\right)^2}=\frac{27}{\left(x+y+z\right)^2}+\frac{\left(x+y+z\right)^4}{27}\)
\(Q\ge\frac{27}{64\left(x+y+z\right)^2}+\frac{27}{64\left(x+y+z\right)^2}+\frac{\left(x+y+z\right)^4}{27}+\frac{837}{32\left(x+y+z\right)^2}\)
\(Q\ge3\sqrt[3]{\frac{27^2\left(x+y+z\right)^4}{64^2.27\left(x+y+z\right)^4}}+\frac{837}{32.\left(\frac{3}{2}\right)^2}=\frac{195}{16}\)
"=" \(\Leftrightarrow x=y=z=\frac{1}{2}\)
Nguyễn Trúc Giang, Duy Khang, Vũ Minh Tuấn, Võ Hồng Phúc, tth, No choice teen, Phạm Lan Hương,
Nguyễn Lê Phước Thịnh, @Nguyễn Việt Lâm, @Akai Haruma
giúp em vs ạ! Cần trước 5h chiều nay ạ
Thanks nhiều
Cho x,y,z >0 thỏa mãn \(\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}>=6\sqrt{z}\) tìm min P =\(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)
cho x,y,z>0 thỏa mãn:\(\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}=2018\)
tìm Min:\(T=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)
Cho x, y, z > 0 thỏa mãn \(\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}=2018\)
Tìm min \(A=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)
Đặt \(\sqrt{x^2+y^2}=c;\sqrt{y^2+z^2}=a;\sqrt{z^2+x^2}=b\)
Ta có:
\(A=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)
\(\ge\frac{x^2}{\sqrt{2\left(y^2+z^2\right)}}+\frac{y^2}{\sqrt{2\left(z^2+x^2\right)}}+\frac{z^2}{\sqrt{2\left(x^2+y^2\right)}}\)
\(=\frac{1}{2\sqrt{2}}\left(\frac{c^2+b^2-a^2}{a}+\frac{a^2+c^2-b^2}{b}+\frac{b^2+a^2-c^2}{c}\right)\)
\(\ge\frac{1}{2\sqrt{2}}\left(\frac{\left(2a+2b+2c\right)^2}{2\left(a+b+c\right)}-2018\right)=\frac{1009}{\sqrt{2}}\)
Cho x,y,z>0 thỏa mãn xyz=1. Tìm min \(P=\frac{x^2\left(y+z\right)}{y\sqrt{y}+2z\sqrt{z}}+\frac{y^2\left(z+x\right)}{z\sqrt{z}+2x\sqrt{x}}+\frac{z^2\left(x+y\right)}{x\sqrt{x}+2y\sqrt{y}}\)
bạn vào trang này nhé có bài như thến này đấy
//123doc.org//document/3173507-ren-luyen-chuyen-de-tim-maxmin-on-thi-thpt-quoc-gia.htm
tính diện tích hình vẽ dưới đây
Với 3 số x, y, z thỏa mãn: \(xy+yz+zx=5\)
Tìm Min P= \(\frac{3x+3y+2z}{\sqrt{6\left(x^2+5\right)}+\sqrt{6\left(y^2+5\right)}+\sqrt{z^2+5}}\)
Từ giả thiết \(xy+yz+zx=5\)
ta có \(x^2+5=x^2+xy+yz+zx=\left(x+y\right)\left(z+x\right)\)
Áp dụng BĐT AM-GM , ta có
\(\sqrt{6\left(x^2+5\right)}=\sqrt{6\left(x+y\right)\left(z+x\right)}\le\frac{3\left(x+y\right)+2\left(z+x\right)}{2}=\frac{5x+3y+2z}{2}\)
CM tương tự ta được \(\sqrt{6\left(y^2+5\right)}\le\frac{3x+5y+2z}{2};\sqrt{z^2+5}\le\frac{x+y+2z}{2}\)
Cộng zế zới zế BĐt trên ta đc
\(\sqrt{6\left(x^2+5\right)}+\sqrt{6\left(y^2+5\right)}+\sqrt{z^2+5}\le\frac{9x+9y+6z}{2}\)
\(=>P=\frac{3x+3y+2z}{\sqrt{6\left(x^2+5\right)}+\sqrt{6\left(y^2+5\right)}+\sqrt{x^2+5}}\ge\frac{2\left(3x+3y+2z\right)}{9x+9y+6z}=\frac{2}{3}\)
=> \(GTNN\left(P\right)=\frac{2}{3}khi\left(x=y=1;z=2\right)\)
Ta có \(\sqrt{6\left(x^2+5\right)}+\sqrt{6\left(y^2+5\right)}+\sqrt{z^2+5}=\sqrt{6\left(x+y\right)\left(x+z\right)}+\sqrt{6\left(y+z\right)\left(y+x\right)}\)\(+\sqrt{6\left(z+x\right)\left(z+y\right)}\)
\(\le\frac{3\left(x+y\right)+2\left(x+z\right)}{2}+\frac{3\left(x+y\right)+2\left(y+z\right)}{2}+\frac{\left(z+x\right)+\left(z+y\right)}{2}\le\frac{9x+9y+6z}{2}=\frac{3}{2}\)\(\left(3x+3y+2z\right)\)
\(\Rightarrow P=\frac{3x+3y+2z}{\sqrt{6\left(x^2+5\right)}+\sqrt{6\left(y^2+5\right)}+\sqrt{z^2+5}}\ge\frac{2}{3}\)
dấu "=" xảy ra \(\Leftrightarrow x=y=1;z=2\)
Vậy \(P_{min}=\frac{2}{3}\Leftrightarrow x=y=1;z=2\)
cho x,y,z dương thỏa mãn \(xy+yz+zx=3\)Tìm min \(P=\frac{x^2}{\sqrt{x^3+8}}+\frac{y^2}{\sqrt{y^3+8}}+\frac{z^2}{\sqrt{z^3+8}}\)
\(\frac{\left(x+y+z\right)^2}{3}\ge xy+yz+zx\Rightarrow x+y+z\ge3\)
\(P=\frac{x^2}{\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}}+\frac{y^2}{\sqrt{\left(y+2\right)\left(y^2-2y+4\right)}}+\frac{z^2}{\sqrt{\left(z+2\right)\left(z^2-2z+4\right)}}\)
\(\Rightarrow P\ge\frac{\left(x+y+z\right)^2}{\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}+\sqrt{\left(y+2\right)\left(y^2-2y+4\right)}+\sqrt{\left(z+2\right)\left(z^2-2z+4\right)}}\)
\(\Rightarrow P\ge\frac{2\left(x+y+z\right)^2}{\left(x+2+x^2-2x+4\right)+\left(y+2+y^2-2y+4\right)+\left(z+2+z^2-2z+4\right)}\)
\(\Rightarrow P\ge\frac{2\left(x+y+z\right)^2}{\left(x^2+y^2+z^2\right)-\left(x+y+z\right)+18}=\frac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2-\left(x+y+z\right)-2\left(xy+yz+zx\right)+18}=\frac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2-\left(x+y+z\right)+12}\)
Dự đoán Min P=1 khi x+y+z=3
Đặt \(t=x+y+z\ge3\)
\(\Rightarrow P\ge\frac{2t^2}{t^2-t+12}\Rightarrow P-1\ge\frac{t^2+t-12}{t^2-t+12}=\frac{\left(t-3\right)\left(t+4\right)}{t^2-t+12}\ge0\)
\(\Rightarrow P\ge1\)