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
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 \(=\frac{3}{2}\) <=> \(x=y=z=\sqrt{3}\)
