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Xét: \(x^4+y^4-xy\left(x^2+y^2\right)=\left(x^2+y^2+xy\right)\left(x-y\right)^2\ge0\)
\(\Rightarrow x^4+y^4\ge xy\left(x^2+y^2\right)\)(*)
Tương tự với (*) ta có: \(\hept{\begin{cases}y^4+z^4\ge yz\left(y^2+z^2\right)\\z^4+x^4\ge zx\left(z^2+x^2\right)\end{cases}}\)
\(\Rightarrow\Sigma_{cyc}\frac{1}{x^4+y^4+z}\le\Sigma_{cyc}\frac{1}{xy\left(x^2+y^2\right)+z.xyz}=\Sigma_{cyc}\frac{1}{xy\left(x^2+y^2+z^2\right)}=\frac{x+y+z}{x^2+y^2+z^2}\)
Ta có:\(x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2\) và \(x+y+z\ge3\sqrt[3]{xyz}=3\)
\(\Rightarrow\Sigma_{cyc}\frac{1}{x^4+y^4+z}\le\frac{x+y+z}{x^2+y^2+z^2}\le\frac{1}{\frac{1}{3}\left(x+y+z\right)}\le1\)
Dấu "=" xảy ra khi x=y=z=1
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Có : \(P=\Sigma\frac{x}{x+1}\)
\(\Rightarrow3-P=\Sigma\left(1-\frac{x}{x+1}\right)\)
\(=\Sigma\frac{1}{x+1}\)
Áp dụng bđt \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\left(a,b,c>0\right)\)được
\(3-P=\Sigma\frac{1}{x+1}\ge\frac{9}{x+y+z+3}=\frac{9}{4}\)
\(\Rightarrow P\le3-\frac{9}{4}=\frac{3}{4}\)
Dấu "=" khi x = y = z = 1/3
Đặt a = x + 1 > 0 ; b = y + 1 > 0 ; c = z + 4 > 0
a + b + c = 6
\(A=\frac{a-1}{a}+\frac{b-1}{b}+\frac{c-4}{c}=3-\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\)
Theo Bất Đẳng Thức ta có: \(\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{4}{c}\ge\frac{4}{a+b}+\frac{4}{c}\ge\frac{16}{a+b+c}=\frac{8}{3}\)
\(\Rightarrow A\le\frac{1}{3}\)Đẳng thức xảy ra khi và chỉ khi \(\hept{\begin{cases}a=b\\a+b=c\\a+b+c=6\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b=\frac{3}{2}\\c=3\end{cases}\Leftrightarrow\hept{\begin{cases}x=y=\frac{1}{2}\\z=-1\end{cases}}}\)
Vậy MaxA = 1/3 khi \(\hept{\begin{cases}x=y=\frac{1}{2}\\z=-1\end{cases}}\)
\(\frac{16}{2x+y+z}=\frac{16}{x+x+y+z}\le\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\)
Tương tự:
\(\frac{16}{x+2y+z}\le\frac{1}{x}+\frac{2}{y}+\frac{1}{z};\frac{16}{x+y+2z}\le\frac{1}{x}+\frac{1}{y}+\frac{2}{z}\)
Cộng lại:
\(16P\le4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=16\Rightarrow P\le1\)
dấu "=" xảy ra tại \(x=y=z=\frac{3}{4}\)
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 \(=\frac{3}{2}\) <=> \(x=y=z=\sqrt{3}\)
\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\Rightarrow2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge4\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)
\(\Leftrightarrow2\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\le1\)
\(\frac{1}{x+y}+\frac{1}{y+z}\ge\frac{4}{2x+y+z}\Rightarrow2\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\ge4\left(\frac{1}{2x+y+z}+\frac{1}{2y+z+x}+\frac{1}{2z+x+y}\right)\)
\(4M\le1\Leftrightarrow M\le\frac{1}{4}\) \(M=\frac{1}{4}\Leftrightarrow x=y=z=3\)
Ta đi c/m BĐT sau: \(x^3+y^3\ge xy\left(x+y\right)\) (*)
Thật vậy (*) \(\Leftrightarrow x^3+y^3-x^2y-xy^2\ge0\)
\(\Leftrightarrow x^2\left(x-y\right)+y^2\left(y-x\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)\left(x^2-y^2\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\ge0\)(luôn đúng)
Áp dụng vào bài toán:
\(\frac{1}{x^3+y^3+1}\le\frac{1}{xy\left(x+y\right)+1}=\frac{1}{xy\left(x+y+z\right)}\)(Do xyz=1)
Tương tự: \(\frac{1}{y^3+z^3+1}\le\frac{1}{yz\left(x+y+z\right)};\frac{1}{z^3+x^3+1}\le\frac{1}{zx\left(x+y+z\right)}\)
\(\Rightarrow A\le\frac{1}{xy\left(x+y+z\right)}+\frac{1}{yz\left(x+y+z\right)}+\frac{1}{zx\left(x+y+z\right)}=\frac{x+y+z}{xyz\left(x+y+z\right)}=1\)
Vậy Max A = 1. Dấu "=" xảy ra <=> x=y=z=1.
\(3-S=1-\frac{x}{x+1}+1-\frac{y}{y+1}+1-\frac{z}{z+1}=\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\)
\(\Rightarrow3-S\ge\frac{9}{x+y+z+3}=\frac{9}{4}\)
\(\Rightarrow S\le3-\frac{9}{4}=\frac{3}{4}\)
\(\Rightarrow S_{max}=\frac{3}{4}\) khi \(x=y=z=\frac{1}{3}\)