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CT
14 tháng 8 2020
áp dụng bunhiacopski ta có:
P^2 =< (1+1+1)(1/1+x^2 + 1/1+y^2+1/1+z^2)= 3(....)
đặt (...) =A
ta có: 1/1+x^2=< 1/2x
tt với 2 cái kia
=> A=< 1/2(1/x+1/y+1/z) =<1/2 ( xy+yz+xz / xyz)=1/2 ..........
đoạn sau chj chịu
^^ sorry
14 tháng 8 2020
Bài này là câu lớp 8 rất quen thuộc rùiiiiiii !!!!!!!!
gt <=> \(\frac{x+y+z}{xyz}=1\)
<=> \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\)
Đặt: \(\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c\)
=> \(ab+bc+ca=1\)
VÀ: \(x=\frac{1}{a};y=\frac{1}{b};z=\frac{1}{c}\)
THAY VÀO P TA ĐƯỢC:
\(P=\frac{1}{\sqrt{1+\frac{1}{a^2}}}+\frac{1}{\sqrt{1+\frac{1}{b^2}}}+\frac{1}{\sqrt{1+\frac{1}{c^2}}}\)
=> \(P=\frac{1}{\sqrt{\frac{a^2+1}{a^2}}}+\frac{1}{\sqrt{\frac{b^2+1}{b^2}}}+\frac{1}{\sqrt{\frac{c^2+1}{c^2}}}\)
=> \(P=\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\)
Thay \(1=ab+bc+ca\) vào P ta sẽ được:
=> \(P=\frac{a}{\sqrt{a^2+ab+bc+ca}}+\frac{b}{\sqrt{b^2+ab+bc+ca}}+\frac{c}{\sqrt{c^2+ab+bc+ca}}\)
=> \(P=\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(b+a\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)
=> \(2P=2.\sqrt{\frac{a}{a+b}}.\sqrt{\frac{a}{a+c}}+2.\sqrt{\frac{b}{b+a}}.\sqrt{\frac{b}{b+c}}+2.\sqrt{\frac{c}{c+a}}.\sqrt{\frac{c}{c+b}}\)
TA ÁP DỤNG BĐT CAUCHY 2 SỐ SẼ ĐƯỢC:
=> \(2P\le\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{b+a}+\frac{b}{b+c}+\frac{c}{c+a}+\frac{c}{c+b}\)
=> \(2P\le\left(\frac{a}{a+b}+\frac{b}{b+a}\right)+\left(\frac{b}{b+c}+\frac{c}{c+b}\right)+\left(\frac{c}{c+a}+\frac{a}{a+c}\right)\)
=> \(2P\le\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\)
=> \(2P\le1+1+1=3\)
=> \(P\le\frac{3}{2}\)
DẤU "=" XẢY RA <=> \(a=b=c\) . MÀ \(ab+bc+ca=1\)
=> \(a=b=c=\sqrt{\frac{1}{3}}\)
=> \(x=y=z=\sqrt{3}\)
VẬY P MAX \(=\frac{3}{2}\) <=> \(x=y=z=\sqrt{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 4 2022
\(x+y+z=xyz\Rightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
Đặt \(\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)=\left(a;b;c\right)\Rightarrow ab+bc+ca=1\)
\(P=\dfrac{2a}{\sqrt{1+a^2}}+\dfrac{b}{\sqrt{1+b^2}}+\dfrac{c}{\sqrt{1+c^2}}=\dfrac{2a}{\sqrt{ab+bc+ca+a^2}}+\dfrac{b}{\sqrt{ab+bc+ca+b^2}}+\dfrac{c}{\sqrt{ab+bc+ca+c^2}}\)
\(P=\dfrac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)
\(P=\sqrt{\dfrac{2a}{a+b}.\dfrac{2a}{a+c}}+\sqrt{\dfrac{2b}{a+b}.\dfrac{b}{2\left(b+c\right)}}+\sqrt{\dfrac{2c}{c+a}.\dfrac{c}{2\left(c+b\right)}}\)
\(P\le\dfrac{1}{2}\left(\dfrac{2a}{a+b}+\dfrac{2a}{a+c}+\dfrac{2b}{a+b}+\dfrac{b}{2\left(b+c\right)}+\dfrac{2c}{c+a}+\dfrac{c}{2\left(c+b\right)}\right)=\dfrac{9}{4}\)
\(P_{max}=\dfrac{9}{4}\) khi \(\left(a;b;c\right)=\left(\dfrac{7}{\sqrt{15}};\dfrac{1}{\sqrt{15}};\dfrac{1}{\sqrt{15}}\right)\) hay \(\left(x;y;z\right)=\left(\dfrac{\sqrt{15}}{7};\sqrt{15};\sqrt{15}\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 12 2020
\(P\le\sqrt{3\left(\sum\dfrac{1}{\left(x+y\right)^2+\left(x+1\right)^2+4}\right)}\le\sqrt{3\left(\sum\dfrac{1}{4xy+4x+4}\right)}\)
\(P\le\sqrt{\dfrac{3}{4}\sum\left(\dfrac{1}{xy+x+1}\right)}=\dfrac{\sqrt{3}}{2}\)
\(P_{max}=\dfrac{\sqrt{3}}{2}\) khi \(x=y=z=1\)
12 tháng 11 2018
\(A=\frac{1}{\sqrt{x^2-xy+y^2}}+\frac{1}{\sqrt{y^2-yz+z^2}}+\frac{1}{\sqrt{z^2-zx+x^2}}\)
\(=\frac{1}{\sqrt{\frac{1}{2}\left(x-y\right)^2+\frac{1}{2}\left(x^2+y^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(y-z\right)^2+\frac{1}{2}\left(y^2+z^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(z-x\right)^2+\frac{1}{2}\left(z^2+x^2\right)}}\)
\(\le\frac{1}{\sqrt{\frac{1}{2}\left(x^2+y^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(y^2+z^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(z^2+x^2\right)}}\)
\(\le\frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{z+x}\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
MD
1
16 tháng 7 2021
\(=>A=\dfrac{\sqrt{x-1}}{x}+\dfrac{\sqrt{y-2}}{y}+\dfrac{\sqrt{z-3}}{z}\)
áp dụng BĐT AM-GM
\(=>\sqrt{x-1}\le\dfrac{x-1+1}{2}=\dfrac{x}{2}\)
\(=>\dfrac{\sqrt{x-1}}{x}\le\dfrac{\dfrac{x}{2}}{x}=\dfrac{1}{2}\left(1\right)\)
có \(\dfrac{\sqrt{y-2}}{y}=\dfrac{\sqrt{\left(y-2\right)2}}{\sqrt{2}.y}\)
\(=>\sqrt{\left(y-2\right)2}\le\dfrac{y-2+2}{2}=\dfrac{y}{2}\)
\(=>\dfrac{\sqrt{\left(y-2\right)2}}{\sqrt{2}.y}\le\dfrac{\dfrac{y}{2}}{\sqrt{2}.y}=\dfrac{1}{2\sqrt{2}}\left(2\right)\)
tương tự \(=>\dfrac{\sqrt{z-3}}{z}\le\dfrac{1}{2\sqrt{3}}\left(3\right)\)
(1)(2)(3)\(=>A\le\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\)