(Lạng Sơn)
Cho \(x,y\) là hai số dương thay đổi luôn thỏa mãn \(x+2y\le3\). Tìm GTLN của biểu thức \(S=\sqrt{x+3}+2\sqrt{y+3}\).
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VT
30 tháng 4 2018
Áp dụng BĐT bu-nhi-a , ta có \(\left(\sqrt{x+3}+2\sqrt{y+3}\right)^2\le\left(1+2\right)\left(x+3+2y+6\right)\le36\)
=> \(S\le6\)
dấu = xảy ra <=> x=y=1
18 tháng 10 2020
Vì xyz=1\(\Rightarrow x^2\left(y+z\right)\ge2x^2\sqrt{yz}=2x\sqrt{x}\)
Tương tự \(y^2\left(z+x\right)\ge2y\sqrt{y};z^2=\left(x+y\right)\ge2z\sqrt{z}\)
\(\Rightarrow P\ge\frac{2x\sqrt{x}}{y\sqrt{y}+2z\sqrt{z}}+\frac{2y\sqrt{y}}{z\sqrt{z}+2x\sqrt{x}}+\frac{2z\sqrt{z}}{x\sqrt{x}+2y\sqrt{y}}\)
Đặt \(x\sqrt{x}+2y\sqrt{y}=a;y\sqrt{y}+2z\sqrt{z}=b;z\sqrt{z}+2x\sqrt{x}=c\)
\(\Rightarrow x\sqrt{x}=\frac{4c+a-2b}{9};y\sqrt{y}=\frac{4a+b-2c}{9};z\sqrt{z}=\frac{4b+c-2a}{9}\)
\(\Rightarrow P\ge\frac{2}{9}\left(\frac{4c+a-2b}{b}+\frac{4a+b-2c}{a}+\frac{4b+c-2a}{b}\right)\)
\(=\frac{2}{9}\text{ }\left[4\left(\frac{c}{b}+\frac{a}{c}+\frac{b}{a}\right)+\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-6\right]\ge\frac{2}{9}\left(4.3+2-6\right)=2\)
Min P =2 khi và chỉ khi a=b=c khi va chỉ khi x=y=z=1
23 tháng 7 2023
a) \(\left\{{}\begin{matrix}a=x\\b=2y\\c=3z\end{matrix}\right.\Rightarrow a+b+c=2;a,b,c>0\)
\(\Rightarrow S=\sqrt{\dfrac{\dfrac{ab}{2}}{\dfrac{ab}{2}+c}}+\sqrt{\dfrac{\dfrac{bc}{2}}{\dfrac{bc}{2}+a}}+\sqrt{\dfrac{ca}{ca+2b}}\)
\(=\sqrt{\dfrac{ab}{ab+2c}}+\sqrt{\dfrac{bc}{bc+2a}}+\sqrt{\dfrac{ca}{ca+2b}}\)
Vì a,b,c>0 nên áp dụng BĐT AM-GM, ta có:
\(\sqrt{\dfrac{ab}{ab+2c}}=\sqrt{\dfrac{ab}{ab+\left(a+b+c\right)c}}=\sqrt{\dfrac{ab}{c^2+bc+ca+ab}}=\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}\)
\(=\sqrt{\dfrac{a}{a+c}}.\sqrt{\dfrac{b}{b+c}}\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{b}{b+c}\right)\)
\(\sqrt{\dfrac{bc}{bc+2a}}=\sqrt{\dfrac{bc}{\left(b+a\right)\left(c+a\right)}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{c}{a+c}\right)\)
\(\sqrt{\dfrac{ca}{ca+2b}}=\sqrt{\dfrac{ca}{\left(c+b\right)\left(a+b\right)}}\le\dfrac{1}{2}\left(\dfrac{c}{b+c}+\dfrac{a}{a+b}\right)\)
\(\Rightarrow S\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}\right)+\dfrac{1}{2}\left(\dfrac{b}{b+c}+\dfrac{c}{b+c}\right)+\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{c}{a+c}\right)=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi và chỉ khi: a=b=c=2/3=>\(\left(x,y,z\right)=\left\{\dfrac{2}{3};\dfrac{1}{3};\dfrac{2}{9}\right\}\)
7 tháng 9 2018
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 \(\Leftrightarrow x=y=z=\frac{1}{3}\)
NM
Nguyễn Minh Quang
Giáo viên
28 tháng 7 2021
Áp dụng bất đẳng thức Bunhia ta có :
\(\left(\sqrt{1+x^2}+\sqrt{2x}\right)^2\le2\left(1+x^2+2x\right)=2\left(x+1\right)^2\text{ nên }\sqrt{1+x^2}+\sqrt{2x}\le\sqrt{2}\left(x+1\right)\)
tương tự ta có : \(\hept{\begin{cases}\sqrt{1+y^2}+\sqrt{2y}\le\sqrt{2}\left(y+1\right)\\\sqrt{1+z^2}+\sqrt{2z}\le\sqrt{2}\left(z+1\right)\end{cases}}\)
Nên \(A\le\sqrt{2}\left(x+y+z+3\right)+\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\left(2-\sqrt{2}\right)\)
\(\le6\sqrt{2}+\left(2-\sqrt{2}\right)\sqrt{3\left(x+y+z\right)}\le6\sqrt{2}+\left(2-\sqrt{2}\right).3=6+3\sqrt{2}\)
dấu bằng xảy ra khi x=y=z=1
16 tháng 6 2019
Ta có \(\left(2x^2+y^2+3\right)\left(2+1+3\right)\ge\left(2x+y+3\right)^2\)
=> \(\frac{1}{\sqrt{2x^2+y^2+3}}\le\frac{\sqrt{6}}{2x+y+3}\)
Mà \(\frac{1}{2x+y+3}=\frac{1}{x+x+y+1+1+1}\le\frac{1}{36}\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+3\right)\)
=> \(\frac{1}{\sqrt{2x^2+y^2+3}}\le\frac{\sqrt{6}}{36}\left(\frac{2}{x}+\frac{1}{y}+3\right)\)
Khi đó
\(P\le\frac{\sqrt{6}}{36}\left(\frac{3}{x}+\frac{3}{y}+\frac{3}{z}+9\right)=\frac{\sqrt{6}}{36}.18=\frac{\sqrt{6}}{2}\)
Dấu bằng xảy ra khi x=y=z=1
Vậy \(MaxP=\frac{\sqrt{6}}{2}\)khi x=y=z=1
KH
1
HP
16 tháng 1 2021
Từ giả thiết ta có:
\(x+y=3\left(\sqrt{x+1}+\sqrt{y+2}\right)\le3\sqrt{2\left(x+y+3\right)}\)
\(\Leftrightarrow P\le3\sqrt{2\left(P+3\right)}\)
\(\Leftrightarrow\left\{{}\begin{matrix}P\ge0\\18P+54\ge P^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}P\ge0\\P^2-18P-54\le0\end{matrix}\right.\)
\(\Leftrightarrow0\le P\le9+3\sqrt{15}\)
\(\Rightarrow maxP=9+3\sqrt{15}\Leftrightarrow\left(x;y\right)=\left(\dfrac{10+3\sqrt{15}}{2};\dfrac{8+3\sqrt{15}}{2}\right)\)