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Đặt: \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{xyz}\)
\(\Leftrightarrow xy+yz+zx=1\)
Ta có:
\(S=\frac{\frac{1}{x}}{\sqrt{\frac{1}{y}.\frac{1}{z}\left(1+\frac{1}{x^2}\right)}}+\frac{\frac{1}{y}}{\sqrt{\frac{1}{z}.\frac{1}{x}\left(1+\frac{1}{y^2}\right)}}+\frac{\frac{1}{z}}{\sqrt{\frac{1}{x}.\frac{1}{y}\left(1+\frac{1}{z^2}\right)}}\)
\(=\sqrt{\frac{yz}{1+x^2}}+\sqrt{\frac{zx}{1+y^2}}+\sqrt{\frac{xy}{1+z^2}}\)
\(=\sqrt{\frac{yz}{xy+yz+zx+x^2}}+\sqrt{\frac{zx}{xy+yz+zx+y^2}}+\sqrt{\frac{xy}{xy+yz+zx+z^2}}\)
\(=\sqrt{\frac{yz}{\left(x+y\right)\left(x+z\right)}}+\sqrt{\frac{zx}{\left(y+x\right)\left(y+z\right)}}+\sqrt{\frac{xy}{\left(z+x\right)\left(z+y\right)}}\)
\(\le\frac{1}{2}.\left(\frac{y}{x+y}+\frac{z}{x+z}+\frac{z}{y+z}+\frac{x}{x+y}+\frac{x}{z+x}+\frac{y}{z+y}\right)\)
\(=\frac{1}{2}.\left(1+1+1\right)=\frac{3}{2}\)
Dấu = xảy ra khi \(x=y=z=\sqrt{3}\)
Vì \(abc=1\)nên trong 3 số a,b,c luôn có 2 số nằm cùng phía so với 1.
Không mất tính tổng quát ta giả sử 2 số đó là a và b, khi đó ta có:
\(\left(1-a\right)\left(1-b\right)\ge0\Leftrightarrow a+b\le1+ab=\frac{c+1}{c}\)
Do đó ta được:
\(\left(a+1\right)\left(b+1\right)\left(c+1\right)=\left(1+a+b+ab\right)\left(c+1\right)\)
\(=2\left(1+ab\right)\left(1+c\right)\le\frac{2\left(c+1\right)^2}{c}\)
Áp dụng bất đẳng thức Bunhiacopxki ta có:
\(\frac{1}{\left(1+a\right)^2}+\frac{1}{\left(1+b\right)^2}\ge\frac{1}{\left(1+ab\right)\left(1+\frac{a}{b}\right)}+\frac{1}{\left(1+ab\right)\left(1+\frac{b}{a}\right)}\)
\(=\frac{b}{\left(1+ab\right)\left(a+b\right)}+\frac{a}{\left(1+ab\right)\left(a+b\right)}=\frac{1}{1+ab}=\frac{c}{c+1}\)
Do đó ta được:
\(\frac{1}{\left(1+a\right)^2}+\frac{1}{\left(1+b\right)^2}+\frac{1}{\left(1+c\right)^2}+\frac{2}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\)
\(\ge\frac{c}{c+1}+\frac{1}{\left(c+1\right)^2}+\frac{c}{\left(c+1\right)^2}=\frac{c\left(c+1\right)+1+c}{\left(c+1\right)^2}=1\)
Như vậy bất đẳng thức ban đầu được chứng minh. Đẳng thức xẩy ra khi \(a=b=c=1\).
c) Có \(P=\frac{ax+b}{x^2+1}=-1+\frac{x^2+ax+b+1}{x^2+1}\);
\(P=\frac{ax+b}{x^2+1}=4-\frac{4x^2-ax-b+4}{x^2+1}\)
Để Min P = 1 và Max P = 4 thì
\(\hept{\begin{cases}x^2+ax+b+1=\left(x+c\right)^2\\4x^2-ax-b+4=\left(2x+d\right)^2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\left(a-2c\right)+\left(b+1-c^2\right)=0\left(1\right)\\x\left(-a-4d\right)+\left(-b+4-d^2\right)=0\left(2\right)\end{cases}}\)
(1) = 0 khi \(\hept{\begin{cases}a=2c\\b=c^2-1\end{cases}}\)(3)
(2) = 0 khi \(\hept{\begin{cases}a=-4d\\b=4-d^2\end{cases}}\)(4)
Từ (3) (4) => d = 1 ; c = -2 ; b = 3 ; a = -4
Vậy \(P=\frac{-4x+3}{x^2+1}\)
ĐK \(x\ge y\)
Đặt \(\sqrt{x+y}=a;\sqrt{x-y}=b\left(a;b\ge0\right)\)
HPT <=> \(\hept{\begin{cases}a^4+b^4=82\\a-2b=1\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(2b+1\right)^4+b^4=82\\a=2b+1\end{cases}}\Leftrightarrow\hept{\begin{cases}17b^4+32b^3+24b^2+8b-81=0\\a=2b+1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}17b^4-17b^3+49^3-49b^2+73b^2-73b+81b-81=0\\a=2b+1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(b-1\right)\left(17b^3+49b^2+73b+81\right)=0\left(1\right)\\a=2b+1\end{cases}}\)
Giải (1) ; kết hợp điều kiện => b = 1
=> Hệ lúc đó trở thành \(\hept{\begin{cases}b=1\\a=2b+1\end{cases}}\Leftrightarrow\hept{\begin{cases}b=1\\a=3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x+y}=3\\\sqrt{x-y}=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=9\\x-y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}2x=10\\x-y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=5\\x-y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=5\\y=4\end{cases}}\)
Vậy hệ có 1 nghiệm duy nhất (x;y) = (5;4)
ap dung bdt \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(\frac{1}{2a+b+c}=\frac{1}{\left(a+b\right)+\left(a+c\right)}\le\frac{1}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\)
\(\Rightarrow P\le\frac{1}{16}\left[\left(\frac{1}{a+b}+\frac{1}{a+c}\right)^2+\left(\frac{1}{a+b}+\frac{1}{b+c}\right)^2+\left(\frac{1}{b+c}+\frac{1}{a+c}^2\right)\right]\)
\(\Rightarrow16P\le\frac{2}{\left(a+b\right)^2}+\frac{2}{\left(b+c\right)^2}+\frac{2}{\left(a+c^2\right)}+\frac{2}{\left(a+b\right)\left(b+c\right)}+\frac{2}{\left(a+b\right)\left(a+c\right)}\)\(+\frac{2}{\left(b+c\right)\left(c+a\right)}\)
ap dung \(x^2+y^2+z^2\ge xy+yz+xz\) voi a+b=x, b+c=y, c+a=z
\(16P\le\frac{4}{\left(a+b\right)^2}+\frac{4}{\left(b+c\right)^2}+\frac{4}{\left(c+a\right)^2}\)
tiếp tục áp dụng bdt ban đầu \(\frac{4}{a+b}\le\frac{1}{a}+\frac{1}{b}\)
\(\Rightarrow\frac{1}{\left(a+b\right)^2}\le4.16.\left(\frac{1}{a}+\frac{1}{b}\right)^2\)
\(\Rightarrow16P\le\frac{1}{4}.16\left[\left(\frac{1}{a}+\frac{1}{b}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2+\left(\frac{1}{c}+\frac{1}{a}\right)^2\right]\)
=\(\frac{1}{4}\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}\right)\)
tiep tuc ap dung bo de thu 2 ta co
\(16P\le\frac{1}{4}.4\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=3\)
\(\Rightarrow p\le\frac{3}{16}\)dau =khi a=b=c=1
Xét biểu thức \(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\)
\(=\frac{\left(a+2\right)\left(b+2\right)+\left(b+2\right)\left(c+2\right)+\left(c+2\right)\left(a+2\right)}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\)
\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{abc+2\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\)
\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{\left(abc+ab+bc+ca\right)+\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\)
\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{4+\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\)(Do \(ab+bc+ca+abc=4\)theo giả thiết)
\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}=1\)(***)
Với x,y dương ta có 2 bất đẳng thức phụ sau:
\(2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)(*)
\(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)(**)
Áp dụng (*) và (**), ta có:
\(\frac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le\frac{1}{a+b+4}=\frac{1}{\left(a+2\right)+\left(b+2\right)}\)
\(\le\frac{1}{4}\left(\frac{1}{a+2}+\frac{1}{b+2}\right)\)(1)
Tương tự ta có: \(\frac{1}{\sqrt{2\left(b^2+c^2\right)}+4}\le\frac{1}{4}\left(\frac{1}{b+2}+\frac{1}{c+2}\right)\)(2)
\(\frac{1}{\sqrt{2\left(c^2+a^2\right)}+4}\le\frac{1}{4}\left(\frac{1}{c+2}+\frac{1}{a+2}\right)\)(3)
Cộng từng vế của các bất đẳng thức (1), (2), (3), ta được:
\(P\le\frac{1}{2}\left(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\right)=\frac{1}{2}\)(theo (***))
Đẳng thức xảy ra khi \(a=b=c\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\Rightarrow\frac{ab+bc+ca}{abc}=\frac{1}{abc}\Rightarrow ab+bc+ca=1\)
Khi đó: \(\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)=\left[ab+bc+ca+a^2\right]\left[ab+bc+ca+b^2\right]\left[ab+bc+ca+c^2\right]\)
\(=\left[a\left(a+b\right)+c\left(a+b\right)\right]\left[b\left(a+b\right)+c\left(a+b\right)\right]\left[b\left(a+c\right)+c\left(a+c\right)\right]\)
\(=\left(a+b\right)^2\left(a+c\right)^2\left(b+c\right)^2\)là số chính phương.